as an answer to the stamp with "the proof is left as an exercise to the reader" i want another stamp reading "the rest of the page is intentionally left blank"
@@simono.899 page intentionally left blank is done to tweak with a layout/allignment. en.wikipedia.org/wiki/Intentionally_blank_page lists other use cases too
Astronomers always dismiss any errors in their arithmetic, measurement or conjectures as, "It's only an order-of-magnitude error... nothing to worry about."
I went about it a different way: Start with a^2 + b^2 = c^2. Subtract b^2 from both sides -> a^2 = c^2 - b^2 Difference of squares -> a^2 = (c+b)(c-b) (c-b) is not zero for any right triangle -> (a^2)/(c-b) = c+b Assuming (c-b) = 1 as in the problem formulation, a^2 - b = c, QED
The triplets that are generated in the second case are all just twice the first ones i.e, first case 3,4,5 then second case 6,8,10 it istrue for all cases
It also includes (8, 15, 17) which is not twice of a different one. Though all even ones will be twice that of a triple with difference of 1, yes Edit: All the cases for b+3 are just thrice the first case though! (From what I can find)
Fun fact: as a teenager, I thought I'd shown (to myself) that the only coprime Pythagorean triples were the ones satisfying c=b+1. Since then, I've never really been looking into the theme again, or seen any counterexamples. However, using the algorithm I came up with today and posted in another comment, I found the triple (8,15,17)! (which can also be found by setting a=8 in the algorithm for c=b+2 that's in the video.) By the way, I just revisited the argument from my youth, and I've realized where the mistake was. My current thought is that in coprime Pythagorean triples, b and c can differ only by 1 or by numbers in the form 2k^2. It's not like I've proven it, but it's my immediate intuition after correcting the argument!
I mean, I can prove that all pythagorean triples which have n>1 (where n = c-b) must have the form n=2^(2*m_0+1)*p_1^(2*m_1)*p_2^(2*m_2)*...*p_j^(2*m_j) where all m_j are non-negative integers and all p_j are odd primes. I do not, however, know how often the converse holds.
@@firebrain2991 By the converse, you mean the existence of a coprime Pythagorean triple satisfying c-b=n, where n is in that form? That problem has infinitely many solutions: Let n=2k^2, pick an integer m>k and apply these well-known formulas: a=2mk, b=m^2-k^2, c=m^2+k^2. This gives a Pythagorean triple with c-b=n, and they're coprime if m and k have different parities (odd/even). By the way, I later posted another comment (with different notation) solving both c-b and c-a, including the converse.
if the hypotenuse has to satisfy the condition that it's one more than the long side, here's my even more improved Pythagorean shortcut. Simply take h=l+1 BOOM done Also both 1² mod 4 and 3² mod 4 are congruent to 1 mod 4, therefore a² with odd a is always in the form 4k+1. Since it's equal to 2b+1, it's trivial to see that b must be even.
The complete reasoning I came up with: We know: a=2n+1 and a²=2b+1 therefore: (2n+1)²=2b+1 4n²+4n+1=2b+1 2n²+2n=b let b(n)=2n²+2n let the difference d between 2 consecutive bs be defined as: d(n)=b(n+1)-b(n) d(n)=2(n+1)²+2n+2-2n²-2n d(n)=2n²+2n+2+2-2n d(n)=4n+4 [you could stop here, and call it a QED by being obvious, but nah, I ain't no Engineer] now let the difference δ between 2 consecutive ds be defined as: δ(n)=d(n+1)-d(n) δ(n)=4(n+1)+4-4n-4 δ(n)=4n+4-4n δ(n)=4 therefore, the difference between 2 consecutive bs increases by 4. ∎
9:05 This property (no perfect square of form 2*odd+1) is usually stated as "3 is not a square residual modulo 4". More in general, you have that neither 2 nor 3 are square residuals modulo 4. To prove it, you just need to compute see that (4n)^2=0 (mod 4) (4n+1)^2=1 (mod 4) (4n+2)^2=0 (mod 4) (4n+3)^2=1 (mod 4) This also immediately implies that if a^2+b^2=c^2 then at least one between a and b is even, since if both a and b are odd then c^2=a^2+b^2=2 (mod 4), which contradicts the fact that 2 is not a square residual modulo 4.
I found the triangles at the beginning very curious and I wanted to do some research. I have coded a small program that tells me all the triangles that satisfy that (c1 = sqrt (a ^ 2 + b ^ 2) = (c2 = a ^ 2 - b). And looking at these triangles I have seen several interesting patterns. 1 0 1 3 4 5 5 12 13 7 24 25 9 40 41 11 60 61 13 84 85 15 112 113 17 144 145 19 180 181 21 220 221 23 264 265 25 312 313 27 364 365 29 420 421 31 480 481 33 544 545 35 612 613 37 684 685 39 760 761 As you can see, the first column that would be the minor side of each triangle, follows the pattern 1 3 5 7 9 and its tens consecutively. The second column is more complicated, at a glance it follows a simple order: 0 0 + 4 = 4 4 + 8 = 12 12 + 12 = 24 24 + 16 = 40 . . . But these numbers can also be defined by the polynomial 2x^2 - 2x x = 1 : 2(1^2) - 2(1) = 0 x = 2 : 2(2^2) - 2(2) = 4 x = 3 : 2(3^2) - 2(3) = 12 x = 4 : 2(4^2)- 2(4) = 24 x = 5 : 2(5^2) - 2(5) = 40 . . . . . See? Very interesting hahaha I loved the video, keep it up!!
Every triple can be found with: a=m^2-n^2 b=2mn c=m^2+n^2 For all m,n. (If you have m,n being relatively prime, with one of them being a multiple of 2 you then find all primative triples; proof of this comes from if they're not relatively prime, then they're a multiple of a relatively prime set, and if exactly one is not a multiple of two, you then will get all 3 terms being even). The case showed at the start of the video is from when m = n+1. Kinda interesting the first image you show at 0:44 skips over the (m=4,n=3) which has 7,24,25. You can find more patterns with this as you go with (8,15,17) which comes from m=4,n=1. This triple finds that 2*(17+15)=8^2. Fwiw: Your last example is with m=3,n=1, or 3^1-1^1 = 8, 2*3*1 = 6, 3^2+1^2 = 10.
Papa,i have found a beautiful way to generate triples whose hypotenuse is one more than one of its side,using triangular numbers, (4T)²+(8T+1)=(4T+1)² Where T is any triangular number,and the (8T+1) is a perfect square.Hope you like it.
This beautiful result can also be seen in terms of Gaussian integers: take any positive integers m, n with m>n, work out (m+ni)^2, and work out its modulus squared, and equate it to the square of the modulus of m+ni. Example: (2+i)^2=3+4i, hence 3^2+4^2=5^2 (the 5 comes from 2^2+1^2). Of course this observation doesn't show you get all Pythagorean triples, it just shows you do get only Pythagorean triples.
It doesn't give all of them unless we cancel common factors, right? How about e.g. (9,12,15)? After all, 15 can't be written as the sum of two squares. Seems to me like you can get the coprime triples times any power of 2.
I find it interesting that pythagorean triples where c=b+1 can be found by using the property of generating squares by adding consecutive odd positive integers starting from 1 (1+3+5+7+9+...). So every odd number is a difference of consecutive squares. When this odd number is itself a square, there is a triple [a, (a²-1)/2, (a²+1)/2] - or if I can do this without my keyboard glitching, [2n+1, ((2n+1)²-1)/2, ((2n+1)²+1)/2] for all n in N.
The normal formulation (which generates all co-prime Pythagorean triples) is (a, b, c)=(st, (s^2-t^2)/2, (s^2+t^2)/2) where s > t and both are odd. Using this it's easy to see that the new formula works only for t=1. Obviously, if you multiply all of a, b, and c by a common multiple you get another Pythagorean Triple which is similar to one of the triangles generated by the formula above. (s,t) = (5,3) => (a,b,c) = (15,8,17) is the simplest triple where t>1.
Hey flammy, there is something that can be also very interesting: the sides of a pythagorian triangle can be obtained by choosing teo numbers x and y. The measure of the sides, then, should be equal to x² - y², 2xy and x² + y². All this can be proven by number theory, or evven with the complex function f(z) = z²
The guy on 0:55 showed a trick or something to find the hypotenuse easily "by squaring the smaller leg and subtract the longer leg and the number u get is the hypotenuse measure" But it's not applicable with taking measure of leg as 2,4 3,6 etc plz explain bro
I actually took your advice of "generalizing the algorithm" to cover all pythagorean triples, as since the hypotenuse can only be bigger than the two sides, all pythagorean triples can be made of the form (a,b,b+n), and came out with a way to find all pythagorean triples (which I'm pretty sure has already been discovered but whatever) So what I did is I then took a^2 + b^2 = b^2+2nb+n^2 which removing b^2 becomes a^2 = 2nb +n^2. Sure at this point you could bring c back in, but you can rewrite this as a^2 - n^2 = 2nb. Which if you look at this in the group of the integers modulo n, you find that a^2-0=0, meaning that a^2 must be divisible by n (note this does not mean *a* must be divisible by n), which of course means you can rewrite a^2 in the form k*n, which is important as k turns out to be very important in the cases where the algorithm produces a non-integer set. So with this knowledge in mind, we produce an algorithm (a^2-n^2)/2n = b, which under correct choices of a and n, produces a pythagorean triple (a, b, b+n) (so long as the outputs are all integers). And we know that if b is an integer, c must be too (since n is an integer), so we need only find out when (a^2-n^2)/2n (given n divides a^2, and n
somewhat interesting might be exploring what the stipulation of n must divide a^2 says about the relation of the 3 sides, and also whether or not this double-counts all pythagorean triples (nevermind, not interesting as it *does* do this, altho I'm probably too tired to prove that rigorously, maybe showing how you might restrict n such that it doesn't could be interesting tho)
OK I did another thing based on this same idea: for all odd primes n, n must divide the lengths of all the sides. proof: if n is prime, and divides a^2, then n must do so twice, thus it divides a, but also it must divide k as a^2=nk. Because it divides k, then it must also divide (k-n), and because it divides k-n, and 2b=k-n (taken from the (k-n)/2=b simplification of (a^2-n^2)/2n=b), which means it divides 2b, and because n is *odd*, if it divides 2b, it divides b. Naturally c follows from the extension that if n divides b, it divides (b+n), which equals c.
Hmmm.... it seems to me like all even pythogarean triples are odd triples multiplied by an the root of an even square number (4 ->2, 36->6, 64->8, 144->12 etc). In your example 6-8-10 is a 3-4-5 (X 2, the root of the square number 4). If you look at a^2 + b^2 = c^2 then the proof of that falls out pretty easy. 2a^2 + 2b^2 = 2c^2, the twos cancel and we are left with what we started. You can even generalize more in that the roots of all square number will generate mutiples of pythagorean triples (but we need even roots to make even pythagorean triples).
Consider a^n+b^n = c^n where c is of the form c=b+1. Using the binomial expansion we find a^n = sum from k=1 to k=n of (n choose k) b^n-k. b^n subtracted from both sides so sum starts at 1. “It is impossible…for any number which is a power greater than the second to be written [in this form] I have a truly marvelous demonstration of this proposition which [this UA-cam comment is too small to contain]." -Fermat
Let's say the sides are a, b and b+l. From PT a²+b²=(b+l)² a²=(b+l)²-b²=(b+l+b)(b+l-b)= l(2b+l). Using this, I came up with two ways of generating Pythagorean triplets long before I watched the video. The first one is pretty much like the Python program: you take an odd number as the shortest side, square it and divide the square by 2, rounding down for the larger leg and up for the hypotenuse. The second one can generate every PT (considering ones with different order of legs different) in exactly one way, and here it is: - Take two coprime numbers - Square them - If one of them is even, divide it by two (it should still be even) and multiply the other one by two - Their geometric mean, which is the product of the original numbers, is one leg, their arithmetic mean, which is their semi-sum, is the hypotenuse, and their semi-difference is the other leg - Multiply all sides by the same integer Here's my proof that all the b's in a first way are even: it is known that all odd squares are of form 4k'+1 (k'=k²+k), so if a²=2b+1, 4k'=2b, and b=2k'
So you have a new way to calculate the hypotenuse...if you already know that the hypotenuse is 1 more than the longer of the two legs of the right-angle triangle. In other words, you have to know the hypotenuse already to calculate it by this trick. If we double the sides of the 3, 4, 5 triangle to 6, 8, 10 it doesn't work, but the standard Pythagorean method does.
It's not a new way to calculate the hypotenuse, and it wasn't ever advertised as such. All it is, it's a simple fun fact that can simplify calculations if you already have some information about the problem from experience
yeah, I discovered this on my own back in 11th grade while I was bored in BC Calc class. (I was trying to come up with some other examples of right triangles with integer lengths other than 3-4-5 and 5-12-13, which felt overused by textbooks.)
@@mastershooter64 the german laws are partly influenced by the german philosophy which include, that the will of freedom shouldn't be charged in longer prisen time
@@mastershooter64 In the german legal system, the right to freedom and the right to seek freedom are treated as two seperate things. When you become a prisoner you lose the right to freedom, but you dont lose the right to seek freedom. As such, escaping itself isnt considered a crime. However any illegal acts done during the escape attempt are considered crimes as normal. So bribing guards, doing damage to your cell or other public property, fighting people who try to contain you, etc are all considered illegal, so you will get extra time for those. In practice this means that trying to escape will eventually almost certainly cause you to get more time in prison, as it is nearly impossible to escape from a prison without commiting a single illegal act in the process (although I have heard of a guy that apperantly managed to pull off a completly crime-free escape, although I dont remeber if he was german or from another country with similar laws)
I was curious and found 3 properties. Say a²+b²=c² and b+k=c where all are natural numbers, ♤: a²/k-k=2b (found algebraically) ♤ implies a>k since 2b>0 (also b+a>c=b+k) ♤ also implies k divides a², so say a²=kd, a=sqrt(kd), and say ♡: k=f²g, then d=gh² since the only way to get a natural number as an output is with a perfect square, so a=fgh where h is arbitrary and f and g satisfy ♡ Lastly through feeding ♡ back into ♤, we get g(h²-f²)=2b, where if g is even, h and f are free, but if g is odd, h and f must match parity. In summary pick a difference k=f²g between b and c, and an a=fgh where h>f and if g is even, h is free, if g is odd, h must match the parity of f. Feeding that back will give values of b and c.
Implications include: there exists infinitely many Pythagorean triplets for any natural value of k There exists a Pythagorean triplet for any value of a>=3 (all odd values implied by k=1, even values implied by k=2) The possible values for h depend solely on the power of 2 in the prime factorisation of k (any if odd, even if even, odd if 0)
I found (and proved) a method to get the finitely many (possibly zero) triples satisfying c=a+n. Method: Pick numbers k and m such that m^2=2n/k. In other words, pick a square number that divides 2n, and find the corresponding factor. Then, the triple is: a=k(m+1) c=a+n b=c-k Note that if we pick m=1, we get the triple (4n, 3n, 5n). For m>=2, we get a
I think the most useable thing from this is a^2=2b+1. Because if you know what the hypotenuse and b is, and they follow the criteria to do this, then you can calculate a. But for normal triangles you wont really need it since you dont know if the hypotenuse is 1 more than b. Atleast thats what i get.
so what I got out after that with 2 minutes and a sheet of paper was that for any odd a (>1 for it to make sense), there is always a pythagorean triple [a, (a²-1)/2, (a²+1)/2], simply derived from the a² = 2b+1, which matches your Python table and feels very nice because, well, just one variable TL;DR: for a = 2k+1 with k€N, [a, (a²-1)/2, (a²+1)/2] is a pythagorean triple
To have a pythagorean triple were all sides are even is straightforward: (a,b,c) is a pythagorean triple iff (2a,2b,2c) is one too. (3,4,5) is the smallest pythagorean triple with positive integers. So (6,8,10) is the smallest with all even.
The only place i find weird is 90-45-45degree triangles. (A^2=0.5)+(A^2=0.5)=1 (A^2=1)+(A^2=1)=2 (A^2=1.5)+(A^2=1.5)=3 (A^2=2)+(A^2=2)=4 (A^2=2.5)+(A^2=2.5)=5 The whole numbers are the hypotenuse You can only construct perfect squares and cubes from this. While A^2+B^2=C^2 can construct rectangles, squares, cubes, and cuboids.
The second property also follows pretty easily from the basics of modular arithmetic - the only squares in Z/4Z are 0 and 1, which implies that the squares of two odd numbers will never add up to a perfect square (as it would have residue two, which is impossible).
Just looking at the video, all the pythagorean triplets he has used are of the form (m, (m^2-1)/2, (m^2+1)/2) where m is odd. I thought that this fact was relatively well known, especially since it was given in my maths textbook from primary school, but I guess not. Even my tution teacher was amazed when i used to generate some triplets on the fly. EDIT : btw, the best pythagroean triple is, with the proof left to the reader, (69, 2380, 2381) EDIT : EDIT : ye, you can get any pythagorean triple with the longest sides apart by k, by just taking the above formula, generating one, and scaling/multiplying it by k
I propose a 2-step master plan! Step 1: Use SOH CAH TOA, to get angle: theta = sin^-1(b/(b+1)) theta = sin^-1(b/(b+2)) Step 2: Graph those in Desmos from 0 to 2pi: y=sin^-1(x/(x+1) ) {0
I found that years ago, you can find infinitely many triples by squaring an odd number and taking the floor and ceiling of half that square, was so cool when I noticed, and of course went through with showing why it’s true, hadn’t thought about it for a while by now
Besides triples generated by multiplication by a whole number scalar are all pythogerean triples a b and (b+1) with b increasing in distance away from a
Papa Flammy; A piece of string 30cm long has its two ends joined together and is stretched by 3 pegs to form a triangle. What is the largest triangular area that can be enclosed by the string?
If hypotenuse is b-1 then why don't you subtract 1 to b? Instead of doing (a^2) - b? It may be a very dumb question lol. Or did you mean that this only works when h-1 = b?
Interesting! I think there would be a way to prove that every c = b + k case will only generate a c = b + 1 triangle k times bigger, using trig expression of the angle to establish similitude. And Legendre symbols are probably not that far, in trying to find a condition on a or b being part of a triple.
@@benjaminbrat3922 No. k has to have the following property: For every prime factor p of k holds: If p=2, then p has to have an odd multiplicity (but 2 does not have to be a prime factor of k). If p!=2, then p has to have an even multiplicity. So k = 1, 2, 8, 9, 18, 25, 32... are possible in primitive Pythagorean triples (primitive means that it isn't the multiple of some other Pythagoran triple), the rest isn't. Also, for every of these k, there are infinitely many primitive Pythagorean triples.
I can't seem to make it work. Perhaps im not using it on the right right triangles. But when a is 15 and b 20, c isn't 21. (The b+1 rule). Am I doing something wrong, or did I find an exception?
That * in the title is the absolute clickbait cheat code What I noticed from the fact at 10:18 that all odd numbers are the smaller side of a pythagorean triple: by multiplying all sides by two we can extend this to all numbers that have any odd prime factors, so all natural numbers (greater than two) expect those that are powers of 2 are the smaller side of any pythagorean triple, (qed stamp here)
The last one will be always an odd-even-odd multiplyed by an even number all terms, because you can rescale it completely by dividing all sides by the maximum even common derivator
I'm mainly interested in the coprime triples, and today I realized this: Let (a,b,c) be a coprime Pythagorean triple, and assume b is the even number. Then, c-b is an odd square number, and c-a is twice a square number! The argument also lets us find these triples. Proof: As someone has pointed out before, the triple can be written as: a = n^2-m^2, b = 2nm, c = n^2+m^2. It follows that c-a = 2m^2 and c-b = (n-m)^2. For the triple to be coprime, n and m must themselves be coprime and have different parity (odd and even). Comparing a = (n-m)(n+m) to b = 2nm, it's not hard to verify that this condition is also sufficient. QED Thus, to find a coprime triple with c-b = k^2, where k is odd, we can just pick the numbers n and m so that they satisfy n-m = k. Or if we want c-a = 2k^2, where k can be even or odd, pick m = k and any n with the other parity.
No, but they do satisfy a=m^2-n^2, b=2mn, c=m^2+n^2, where m & n are positive integers with m>n. The cases considered here are where m-n=1. You get a different series for each different value of m-n (if you fix m-n and vary m). m-n=2 was the case considered at the end.
I mean of course 6 8 10 differ by 2, you just take the 3 4 5 perfect square and multiply everything by 2. You can multiply it by any positive integer n you like to get a perfect square with difference n between the numbers... because if a^2 + b^2 = c^2, surely (an)^2 + (bn)^2 = (cn)^2. Still a cool video though :) Another related cool property is that an odd prime p is a perfect square if and only if it is of the form p = 4k + 1, and then only in one way (so there is precisely one a and one b such that a^2 + b^2 = p^2)
It's not really surprising, that the subset of the pythagorean triples you are looking at, doesn't contain a triple for which the side length of a and b are both odd. There is simply no pythagorean triple for which sides a and b could be both odd: All pythagorean triples are of the form (u^2-v^2, 2uv, u^2+v^2), u, v in ℕ, u > v. 2uv is never odd.
I have been doing this for years before and during my undergrad, I though this was a widely known proof!? The fact a^2=b+c was super helpful during a few modules and proof exercises. Also that Even Pythagorean Triples are not unique in angle/angle/angle measures.
I haven't watched quite yet, Jens, but I can just say that I hope this is a video addressing the metric tensor!!! I'm literally writing my own script on a derivation of its properties to give to the students in GR for my university (RPI). You can derive tensor transformation laws from the more fundamental, tensorial quality of multilinearity. Absolutely brilliant! And, the metric, a tensor field describing the entirety of information regarding length, angle, and curvature {all the intrinsic properties} of a metric space (often just called a manifold) as proven by Gauss's theorema egregium {which uses the connection coefficients [Christoffel Symbols] !!}, is so integral to the nature of reality. I like to call the evaluation of the metric on two vectors in the tangent bundle equating to the first fundamental form: the generalized Pythagorean theorem. Lmao I am nerding out here. Sorry LOL!!! I love you so much, by the way! Thank you for the content and for inspiring me to stay excited and in anticipation!
I've spent the day messing around with cases of this and I'm closing in on a formula for triples of the form (a, b, b+2n) for the natural number n. There's a ton of constraints on it but I'm enjoying the process either way, love ur vids bb
Interesting relationships. a^2-b = c where c=b+1 because for any right triangle, (a^2/(c-b) -b = c. Where c-b=1, a^2/(c-b) = a^2. Care to prove (a^2/(c-b))-b=c for any a^2+b^2=c^2? :-)
Something I found: if c=b+1, then c-a is two times a square. Another fact: If the exponent of an odd prime p in c-b is odd, then p divides both a b and c. It also works if we replace c-b by c-a.
didn't exactly follow the video... but does the longest side have to be 1 less than the hypotenuse to work? if not, 6 8 10 doesn't follow this. 36-8= 18 ??
Some random manipulations, with a < b < c three sides of a right triangle (c: hypothenuse): c == b+n --> a^2+b^2 == c^2 == b^2+2bn+n^2 a^2 == (2b+n)n Case (1) n odd : n=2k Case (2) n even: n=2k+1 (1): a^2 == 4(b+k)k --> (a/2)^2 == (b+k)k (2): a^2 == (2(b+k)+1)(2k+1) --> (a^2-1)/2 == bk+(2k+b)(k+1) ---- With b == a+n , get: c^2 == a^2+b^2 == a^2+(a+n)^2 == 2a(a+n)+n^2 == 2ab+n^2
The first one came from considering when the difference between consecutive square numbers is itself square (I.e. (n+1)^2 - n^2 = k^2). The second came from considering the difference between squares separated by one values (I.e. (n+2)^2 - n^2 = k^2) You can carry on like this (so (n+3)^2-n^2 etc), but the conditions you have to place on k to get integer answers get more complicated
Ah, but a (6,8,10) right triangle is really just a disguised (3,4,5) right triangle! You can take any of your odd ones and get even ones by multiplying everything by 2! Because of the properties of similar triangles, scaling all side lengths by the same amount preserves angles.
I once wrote ''proof left to the examiner'' in my linear algebra exam. No regrets.
xD
Lmaooooo
That's how real men do proofs
@@Linoosethemooss you are from his new short based on this?
The proof is trivial and is left as an exercise to the grader
Me in exam:
Multiply both sides by 0
LHS = RHS
Teacher:
Multiplied my marks by 0
HAHAHAHA
Well shouldn't it work
as an answer to the stamp with
"the proof is left as an exercise to the reader"
i want another stamp reading
"the rest of the page is intentionally left blank"
Yeahhhhh.... an approximation meme
What is the origin of this sentence?
@@simono.899 page intentionally left blank is done to tweak with a layout/allignment.
en.wikipedia.org/wiki/Intentionally_blank_page lists other use cases too
@@sr6550 thx mate
"the proof is too large to fit in the margin"
When it Starts with the good old approximation meme.... You know it's gonna be gud
No you know its gonna be trash
Why is 1.000 = 9.999? I just don’t get the meme
@@Firefly256 because for astronomers 1000 m (it was 1 x 10^3) is almost nothing, since they are dealing with gigantic measures
@@fredesch3158 so 1 and 10 are totally the same to them?
@@Firefly256 Do you know how approximation works? Or better, do you know how numbers work in general?
Im taking a stellar evolution class and i can confirm, 10^6 is = 9.9x10^6
:'D
@@PapaFlammy69 Math is love math is live y=z
if y is 2 and z is 3 then 2=3
Ah yes, the astronomer
Astrofizics
Astronomers always dismiss any errors in their arithmetic, measurement or conjectures as, "It's only an order-of-magnitude error... nothing to worry about."
Your chalkboard is trying to escape.
I went about it a different way:
Start with a^2 + b^2 = c^2.
Subtract b^2 from both sides -> a^2 = c^2 - b^2
Difference of squares -> a^2 = (c+b)(c-b)
(c-b) is not zero for any right triangle -> (a^2)/(c-b) = c+b
Assuming (c-b) = 1 as in the problem formulation, a^2 - b = c, QED
I find something very funny about you constantly keeping your board from going up
0:09
Top 10 greatest voice cracks of all times
1.
The triplets that are generated in the second case are all just twice the first ones i.e, first case 3,4,5 then second case 6,8,10 it istrue for all cases
Yes brother
It also includes (8, 15, 17) which is not twice of a different one. Though all even ones will be twice that of a triple with difference of 1, yes
Edit: All the cases for b+3 are just thrice the first case though! (From what I can find)
Everyone : Pythagoras theorem
Some random guy on UA-cam : GoUgu thEOrEm
Me : confused
Hotel : trivago
...
xD
thats presh
12:15 (1, 0, 1) is a Pythagorean triplet 😎
A degenerate One, like (0,0,0), or anything equal to ABC=0.
That doesn't form a triangle.
@@dhanushkumar8188 but is still a Pythagorean triplet
@@dhanushkumar8188, it's a degenerate triangle; one of the "sides" is zero-dimensional! 😋
Da ya da Suri?
If you ever get a chance, watch him after he drinks a redbull!
there's nothing better than post-graduate me.. ah math
I’m sure he hates redbull
Haven't really explored math in a while; thanks for having me revisit it again. 😊
Fun fact: as a teenager, I thought I'd shown (to myself) that the only coprime Pythagorean triples were the ones satisfying c=b+1. Since then, I've never really been looking into the theme again, or seen any counterexamples. However, using the algorithm I came up with today and posted in another comment, I found the triple (8,15,17)! (which can also be found by setting a=8 in the algorithm for c=b+2 that's in the video.)
By the way, I just revisited the argument from my youth, and I've realized where the mistake was. My current thought is that in coprime Pythagorean triples, b and c can differ only by 1 or by numbers in the form 2k^2. It's not like I've proven it, but it's my immediate intuition after correcting the argument!
I mean, I can prove that all pythagorean triples which have n>1 (where n = c-b) must have the form n=2^(2*m_0+1)*p_1^(2*m_1)*p_2^(2*m_2)*...*p_j^(2*m_j) where all m_j are non-negative integers and all p_j are odd primes. I do not, however, know how often the converse holds.
@@firebrain2991 By the converse, you mean the existence of a coprime Pythagorean triple satisfying c-b=n, where n is in that form? That problem has infinitely many solutions: Let n=2k^2, pick an integer m>k and apply these well-known formulas:
a=2mk, b=m^2-k^2, c=m^2+k^2.
This gives a Pythagorean triple with c-b=n, and they're coprime if m and k have different parities (odd/even).
By the way, I later posted another comment (with different notation) solving both c-b and c-a, including the converse.
if the hypotenuse has to satisfy the condition that it's one more than the long side, here's my even more improved Pythagorean shortcut. Simply take h=l+1 BOOM done
Also both 1² mod 4 and 3² mod 4 are congruent to 1 mod 4, therefore a² with odd a is always in the form 4k+1. Since it's equal to 2b+1, it's trivial to see that b must be even.
5:00 A fact I find kinda interesting: the differences between the bs are going up by 4
oh! Didn't notice that! :D
but it does make sense, since b=2n :)
The complete reasoning I came up with:
We know: a=2n+1 and a²=2b+1
therefore: (2n+1)²=2b+1
4n²+4n+1=2b+1
2n²+2n=b
let b(n)=2n²+2n
let the difference d between 2 consecutive bs be defined as:
d(n)=b(n+1)-b(n)
d(n)=2(n+1)²+2n+2-2n²-2n
d(n)=2n²+2n+2+2-2n
d(n)=4n+4
[you could stop here, and call it a QED by being obvious, but nah, I ain't no Engineer]
now let the difference δ between 2 consecutive ds be defined as:
δ(n)=d(n+1)-d(n)
δ(n)=4(n+1)+4-4n-4
δ(n)=4n+4-4n
δ(n)=4
therefore, the difference between 2 consecutive bs increases by 4. ∎
@@janus3042 at the QED point just factor out the 4 and by the assumption that n is an integer u can declare that it is in fact 4
10:18 if you notice every triple that starts with a prime number fulfils the c=b+1 rule since every prime (except 2) is odd!
9:05 This property (no perfect square of form 2*odd+1) is usually stated as "3 is not a square residual modulo 4".
More in general, you have that neither 2 nor 3 are square residuals modulo 4.
To prove it, you just need to compute see that
(4n)^2=0 (mod 4)
(4n+1)^2=1 (mod 4)
(4n+2)^2=0 (mod 4)
(4n+3)^2=1 (mod 4)
This also immediately implies that if a^2+b^2=c^2 then at least one between a and b is even, since if both a and b are odd then c^2=a^2+b^2=2 (mod 4), which contradicts the fact that 2 is not a square residual modulo 4.
I found the triangles at the beginning very curious and I wanted to do some research. I have coded a small program that tells me all the triangles that satisfy that (c1 = sqrt (a ^ 2 + b ^ 2) = (c2 = a ^ 2 - b).
And looking at these triangles I have seen several interesting patterns.
1 0 1
3 4 5
5 12 13
7 24 25
9 40 41
11 60 61
13 84 85
15 112 113
17 144 145
19 180 181
21 220 221
23 264 265
25 312 313
27 364 365
29 420 421
31 480 481
33 544 545
35 612 613
37 684 685
39 760 761
As you can see, the first column that would be the minor side of each triangle, follows the pattern 1 3 5 7 9 and its tens consecutively. The second column is more complicated, at a glance it follows a simple order:
0
0 + 4 = 4
4 + 8 = 12
12 + 12 = 24
24 + 16 = 40
.
.
.
But these numbers can also be defined by the polynomial 2x^2 - 2x
x = 1 : 2(1^2) - 2(1) = 0
x = 2 : 2(2^2) - 2(2) = 4
x = 3 : 2(3^2) - 2(3) = 12
x = 4 : 2(4^2)- 2(4) = 24
x = 5 : 2(5^2) - 2(5) = 40
.
.
.
.
.
See? Very interesting hahaha
I loved the video, keep it up!!
Every triple can be found with:
a=m^2-n^2
b=2mn
c=m^2+n^2
For all m,n. (If you have m,n being relatively prime, with one of them being a multiple of 2 you then find all primative triples; proof of this comes from if they're not relatively prime, then they're a multiple of a relatively prime set, and if exactly one is not a multiple of two, you then will get all 3 terms being even).
The case showed at the start of the video is from when m = n+1. Kinda interesting the first image you show at 0:44 skips over the (m=4,n=3) which has 7,24,25.
You can find more patterns with this as you go with (8,15,17) which comes from m=4,n=1. This triple finds that 2*(17+15)=8^2.
Fwiw: Your last example is with m=3,n=1, or 3^1-1^1 = 8, 2*3*1 = 6, 3^2+1^2 = 10.
that sliding entrance is on another level
:D
Papa,i have found a beautiful way to generate triples whose hypotenuse is one more than one of its side,using triangular numbers,
(4T)²+(8T+1)=(4T+1)²
Where T is any triangular number,and the (8T+1) is a perfect square.Hope you like it.
nice!!!
Amazing Trick...It saves us, like in Competitive exams.
NOTE: It is applicable only when Hypo. is greater than longer side by one unit.
I know a lot of people don't like proofs, but I think they are fun. Make you think outside the box and you realize how cool math is.
Lololol. I agree. But I don't think the people who don't like proofs think math is anything close to cool in the first place!
No i like proofs especially those stupid ones
New strain of Pythagorean Theorem.
5:13 in fact the sides a,b,c can be always expressed as follows: c=m^2 + n^2; b=2mn; a=m^2 - n^2 for some natural numbers m>n
sure, by Euclid! :)
Those expressions seem very similar to rational parametrization of unit circle.
That's a very nice property, thank you.
This beautiful result can also be seen in terms of Gaussian integers: take any positive integers m, n with m>n, work out (m+ni)^2, and work out its modulus squared, and equate it to the square of the modulus of m+ni. Example: (2+i)^2=3+4i, hence 3^2+4^2=5^2 (the 5 comes from 2^2+1^2).
Of course this observation doesn't show you get all Pythagorean triples, it just shows you do get only Pythagorean triples.
It doesn't give all of them unless we cancel common factors, right? How about e.g. (9,12,15)? After all, 15 can't be written as the sum of two squares. Seems to me like you can get the coprime triples times any power of 2.
How to find an even Pythagorean triple: Let (a, b, c) be a Pythagorean triple, then (2a, 2b, 2c) is an even Pythagorean triple, lol.
I find it interesting that pythagorean triples where c=b+1 can be found by using the property of generating squares by adding consecutive odd positive integers starting from 1 (1+3+5+7+9+...). So every odd number is a difference of consecutive squares. When this odd number is itself a square, there is a triple [a, (a²-1)/2, (a²+1)/2] - or if I can do this without my keyboard glitching,
[2n+1, ((2n+1)²-1)/2, ((2n+1)²+1)/2]
for all n in N.
The normal formulation (which generates all co-prime Pythagorean triples) is (a, b, c)=(st, (s^2-t^2)/2, (s^2+t^2)/2) where s > t and both are odd.
Using this it's easy to see that the new formula works only for t=1.
Obviously, if you multiply all of a, b, and c by a common multiple you get another Pythagorean Triple which is similar to one of the triangles generated by the formula above.
(s,t) = (5,3) => (a,b,c) = (15,8,17) is the simplest triple where t>1.
Hey flammy, there is something that can be also very interesting: the sides of a pythagorian triangle can be obtained by choosing teo numbers x and y. The measure of the sides, then, should be equal to x² - y², 2xy and x² + y². All this can be proven by number theory, or evven with the complex function f(z) = z²
ye, that is due to Euclid^^
You can also prove it with geometry too!
Chris Smith sounds like someone who would travel with a time lord in a blue box
You get all the solutions with c=b+2 by multiplying the ones with c=b+1 by 2 and wise versa.
Congratulations for 200k. A poll with 17 options is the perfect way to celebrate.
:DDD
2:58, binomial theorum ? overkill?
The guy on 0:55 showed a trick or something to find the hypotenuse easily "by squaring the smaller leg and subtract the longer leg and the number u get is the hypotenuse measure" But it's not applicable with taking measure of leg as 2,4 3,6 etc plz explain bro
I actually took your advice of "generalizing the algorithm" to cover all pythagorean triples, as since the hypotenuse can only be bigger than the two sides, all pythagorean triples can be made of the form (a,b,b+n), and came out with a way to find all pythagorean triples (which I'm pretty sure has already been discovered but whatever)
So what I did is I then took a^2 + b^2 = b^2+2nb+n^2 which removing b^2 becomes a^2 = 2nb +n^2. Sure at this point you could bring c back in, but you can rewrite this as a^2 - n^2 = 2nb. Which if you look at this in the group of the integers modulo n, you find that a^2-0=0, meaning that a^2 must be divisible by n (note this does not mean *a* must be divisible by n), which of course means you can rewrite a^2 in the form k*n, which is important as k turns out to be very important in the cases where the algorithm produces a non-integer set.
So with this knowledge in mind, we produce an algorithm (a^2-n^2)/2n = b, which under correct choices of a and n, produces a pythagorean triple (a, b, b+n) (so long as the outputs are all integers). And we know that if b is an integer, c must be too (since n is an integer), so we need only find out when (a^2-n^2)/2n (given n divides a^2, and n
somewhat interesting might be exploring what the stipulation of n must divide a^2 says about the relation of the 3 sides, and also whether or not this double-counts all pythagorean triples (nevermind, not interesting as it *does* do this, altho I'm probably too tired to prove that rigorously, maybe showing how you might restrict n such that it doesn't could be interesting tho)
OK I did another thing based on this same idea: for all odd primes n, n must divide the lengths of all the sides.
proof: if n is prime, and divides a^2, then n must do so twice, thus it divides a, but also it must divide k as a^2=nk. Because it divides k, then it must also divide (k-n), and because it divides k-n, and 2b=k-n (taken from the (k-n)/2=b simplification of (a^2-n^2)/2n=b), which means it divides 2b, and because n is *odd*, if it divides 2b, it divides b. Naturally c follows from the extension that if n divides b, it divides (b+n), which equals c.
@@firebrain2991 my man you actually typed that? I feel bad if i don't read it
Can I apply this property everywhere ? Does this work for all triangles ?
Pappa please where did you get that beautiful fern fractal in background??
On STEMerch, my online shop! stemerch.com/collections/posters-wall-decor
Hmmm.... it seems to me like all even pythogarean triples are odd triples multiplied by an the root of an even square number (4 ->2, 36->6, 64->8, 144->12 etc). In your example 6-8-10 is a 3-4-5 (X 2, the root of the square number 4). If you look at a^2 + b^2 = c^2 then the proof of that falls out pretty easy. 2a^2 + 2b^2 = 2c^2, the twos cancel and we are left with what we started. You can even generalize more in that the roots of all square number will generate mutiples of pythagorean triples (but we need even roots to make even pythagorean triples).
Consider a^n+b^n = c^n where c is of the form c=b+1. Using the binomial expansion we find a^n = sum from k=1 to k=n of (n choose k) b^n-k. b^n subtracted from both sides so sum starts at 1. “It is impossible…for any number which is a power greater than the second to be written [in this form] I have a truly marvelous demonstration of this proposition which [this UA-cam comment is too small to contain]." -Fermat
Let's say the sides are a, b and b+l. From PT a²+b²=(b+l)² a²=(b+l)²-b²=(b+l+b)(b+l-b)= l(2b+l).
Using this, I came up with two ways of generating Pythagorean triplets long before I watched the video. The first one is pretty much like the Python program: you take an odd number as the shortest side, square it and divide the square by 2, rounding down for the larger leg and up for the hypotenuse. The second one can generate every PT (considering ones with different order of legs different) in exactly one way, and here it is:
- Take two coprime numbers
- Square them
- If one of them is even, divide it by two (it should still be even) and multiply the other one by two
- Their geometric mean, which is the product of the original numbers, is one leg, their arithmetic mean, which is their semi-sum, is the hypotenuse, and their semi-difference is the other leg
- Multiply all sides by the same integer
Here's my proof that all the b's in a first way are even: it is known that all odd squares are of form 4k'+1 (k'=k²+k), so if a²=2b+1, 4k'=2b, and b=2k'
So you have a new way to calculate the hypotenuse...if you already know that the hypotenuse is 1 more than the longer of the two legs of the right-angle triangle. In other words, you have to know the hypotenuse already to calculate it by this trick. If we double the sides of the 3, 4, 5 triangle to 6, 8, 10 it doesn't work, but the standard Pythagorean method does.
It's not a new way to calculate the hypotenuse, and it wasn't ever advertised as such. All it is, it's a simple fun fact that can simplify calculations if you already have some information about the problem from experience
Love that flying 3b1b!
hehe :3
Just got a bunch of a level pythagoras hw so was excited when I saw this
yeah, I discovered this on my own back in 11th grade while I was bored in BC Calc class. (I was trying to come up with some other examples of right triangles with integer lengths other than 3-4-5 and 5-12-13, which felt overused by textbooks.)
hey papa flammy I heard that escaping from prison in german is actually legal, is that true?
Escaping itself is not illegal, but if the police catches you again, you have to go back, but you won't get an extra punishment for the escaping
@@leonblattmann1118 that's weird, I wonder why the act of escaping a prison isn't illegal in germany
@@mastershooter64 the german laws are partly influenced by the german philosophy which include, that the will of freedom shouldn't be charged in longer prisen time
Though as a foreigner it doesn't exactly make sense it's definitely quite humane to do that.
@@mastershooter64 In the german legal system, the right to freedom and the right to seek freedom are treated as two seperate things. When you become a prisoner you lose the right to freedom, but you dont lose the right to seek freedom. As such, escaping itself isnt considered a crime.
However any illegal acts done during the escape attempt are considered crimes as normal. So bribing guards, doing damage to your cell or other public property, fighting people who try to contain you, etc are all considered illegal, so you will get extra time for those. In practice this means that trying to escape will eventually almost certainly cause you to get more time in prison, as it is nearly impossible to escape from a prison without commiting a single illegal act in the process (although I have heard of a guy that apperantly managed to pull off a completly crime-free escape, although I dont remeber if he was german or from another country with similar laws)
9.999*10^3 and 1.000*10^4
astrophysicists: these are an ORDER OF MAGNITUDE apart, couldn't be any more different
I was curious and found 3 properties.
Say a²+b²=c² and b+k=c where all are natural numbers,
♤: a²/k-k=2b (found algebraically)
♤ implies a>k since 2b>0 (also b+a>c=b+k)
♤ also implies k divides a², so say a²=kd, a=sqrt(kd), and say ♡: k=f²g, then d=gh² since the only way to get a natural number as an output is with a perfect square, so a=fgh where h is arbitrary and f and g satisfy ♡
Lastly through feeding ♡ back into ♤, we get g(h²-f²)=2b, where if g is even, h and f are free, but if g is odd, h and f must match parity.
In summary pick a difference k=f²g between b and c, and an a=fgh where h>f and if g is even, h is free, if g is odd, h must match the parity of f. Feeding that back will give values of b and c.
Implications include:
there exists infinitely many Pythagorean triplets for any natural value of k
There exists a Pythagorean triplet for any value of a>=3 (all odd values implied by k=1, even values implied by k=2)
The possible values for h depend solely on the power of 2 in the prime factorisation of k (any if odd, even if even, odd if 0)
I found (and proved) a method to get the finitely many (possibly zero) triples satisfying c=a+n.
Method: Pick numbers k and m such that m^2=2n/k. In other words, pick a square number that divides 2n, and find the corresponding factor. Then, the triple is:
a=k(m+1)
c=a+n
b=c-k
Note that if we pick m=1, we get the triple (4n, 3n, 5n). For m>=2, we get a
1:23 how dare he not post in LaTeX
That board constantly going up is like having a dog that always tugs on the leash.
xD
Wait a minute where did 2b come from and where did you get all numbers in 4:58, a bit random and no pattern, 5+8=13 and then 13+8=21 not 25, hmmm
I think the most useable thing from this is a^2=2b+1. Because if you know what the hypotenuse and b is, and they follow the criteria to do this, then you can calculate a. But for normal triangles you wont really need it since you dont know if the hypotenuse is 1 more than b. Atleast thats what i get.
Is b always a multiple of 4 or does it just look like that?
how you got all clean chalk board all the time mine is always dusty whatever i do .
so what I got out after that with 2 minutes and a sheet of paper was that for any odd a (>1 for it to make sense), there is always a pythagorean triple [a, (a²-1)/2, (a²+1)/2], simply derived from the a² = 2b+1, which matches your Python table and feels very nice because, well, just one variable
TL;DR: for a = 2k+1 with k€N, [a, (a²-1)/2, (a²+1)/2] is a pythagorean triple
awesome:)
0:40 actually ive known this before but never used it on my schoolworks and exams because you know what they said "because the book says that"
To have a pythagorean triple were all sides are even is straightforward: (a,b,c) is a pythagorean triple iff (2a,2b,2c) is one too.
(3,4,5) is the smallest pythagorean triple with positive integers. So (6,8,10) is the smallest with all even.
The only place i find weird is 90-45-45degree triangles.
(A^2=0.5)+(A^2=0.5)=1
(A^2=1)+(A^2=1)=2
(A^2=1.5)+(A^2=1.5)=3
(A^2=2)+(A^2=2)=4
(A^2=2.5)+(A^2=2.5)=5
The whole numbers are the hypotenuse
You can only construct perfect squares and cubes from this.
While A^2+B^2=C^2 can construct rectangles, squares, cubes, and cuboids.
... yeah, but does it work with 5, 5xsqrt(3), and 10 as the sides?
The second property also follows pretty easily from the basics of modular arithmetic - the only squares in Z/4Z are 0 and 1, which implies that the squares of two odd numbers will never add up to a perfect square (as it would have residue two, which is impossible).
Just looking at the video, all the pythagorean triplets he has used are of the form (m, (m^2-1)/2, (m^2+1)/2) where m is odd. I thought that this fact was relatively well known, especially since it was given in my maths textbook from primary school, but I guess not. Even my tution teacher was amazed when i used to generate some triplets on the fly.
EDIT : btw, the best pythagroean triple is, with the proof left to the reader, (69, 2380, 2381)
EDIT : EDIT : ye, you can get any pythagorean triple with the longest sides apart by k, by just taking the above formula, generating one, and scaling/multiplying it by k
I propose a 2-step master plan!
Step 1:
Use SOH CAH TOA, to get angle:
theta = sin^-1(b/(b+1))
theta = sin^-1(b/(b+2))
Step 2:
Graph those in Desmos from 0 to 2pi:
y=sin^-1(x/(x+1) ) {0
I found that years ago, you can find infinitely many triples by squaring an odd number and taking the floor and ceiling of half that square, was so cool when I noticed, and of course went through with showing why it’s true, hadn’t thought about it for a while by now
papa: the longest side has to be one unit more than---
me: shhhh. now we know:
distance = sqrt(a^2 - b - c - . . .)
Besides triples generated by multiplication by a whole number scalar are all pythogerean triples a b and (b+1) with b increasing in distance away from a
Still waiting for derivation of equation of motion for octuple pendulum with air resistance in cartesian coordinates!
xD In curved space time :DDD
Papa Flammy; A piece of string 30cm long has its two ends joined together and is stretched by 3 pegs to form a triangle. What is the largest triangular area that can be enclosed by the string?
If hypotenuse is b-1 then why don't you subtract 1 to b? Instead of doing (a^2) - b?
It may be a very dumb question lol.
Or did you mean that this only works when h-1 = b?
Interesting! I think there would be a way to prove that every c = b + k case will only generate a c = b + 1 triangle k times bigger, using trig expression of the angle to establish similitude. And Legendre symbols are probably not that far, in trying to find a condition on a or b being part of a triple.
I think my example a=69, b=260, c=269 proves your assumption wrong as a, b and c don't have common factors and c=b+9.
@@BK-nz2vu Nice! Does that work for every k?
@@benjaminbrat3922 No. k has to have the following property: For every prime factor p of k holds: If p=2, then p has to have an odd multiplicity (but 2 does not have to be a prime factor of k). If p!=2, then p has to have an even multiplicity.
So k = 1, 2, 8, 9, 18, 25, 32... are possible in primitive Pythagorean triples (primitive means that it isn't the multiple of some other Pythagoran triple), the rest isn't. Also, for every of these k, there are infinitely many primitive Pythagorean triples.
I can't seem to make it work. Perhaps im not using it on the right right triangles. But when a is 15 and b 20, c isn't 21. (The b+1 rule). Am I doing something wrong, or did I find an exception?
b=(a^2-1)/2 => b=224/2=112 => c=113
That * in the title is the absolute clickbait cheat code
What I noticed from the fact at 10:18 that all odd numbers are the smaller side of a pythagorean triple: by multiplying all sides by two we can extend this to all numbers that have any odd prime factors, so all natural numbers (greater than two) expect those that are powers of 2 are the smaller side of any pythagorean triple, (qed stamp here)
The last one will be always an odd-even-odd multiplyed by an even number all terms, because you can rescale it completely by dividing all sides by the maximum even common derivator
Oh man that thick Scottish (possibly Irish?) accent is beautiful
Pin this.
It's definitely Scottish! Thanks very much :)
Please search pure waah
german
At what point does this break though?!
I'm mainly interested in the coprime triples, and today I realized this:
Let (a,b,c) be a coprime Pythagorean triple, and assume b is the even number. Then, c-b is an odd square number, and c-a is twice a square number! The argument also lets us find these triples.
Proof: As someone has pointed out before, the triple can be written as:
a = n^2-m^2, b = 2nm, c = n^2+m^2.
It follows that c-a = 2m^2 and c-b = (n-m)^2.
For the triple to be coprime, n and m must themselves be coprime and have different parity (odd and even). Comparing a = (n-m)(n+m) to b = 2nm, it's not hard to verify that this condition is also sufficient.
QED
Thus, to find a coprime triple with c-b = k^2, where k is odd, we can just pick the numbers n and m so that they satisfy n-m = k. Or if we want c-a = 2k^2, where k can be even or odd, pick m = k and any n with the other parity.
Do all Pythagorean triple numbers follow, the side length a, b, b+1 configuration where b+1 is the hypotenuse?
No, for example 8,15,17 doesnt satisfy that
No, but they do satisfy a=m^2-n^2, b=2mn, c=m^2+n^2, where m & n are positive integers with m>n. The cases considered here are where m-n=1. You get a different series for each different value of m-n (if you fix m-n and vary m). m-n=2 was the case considered at the end.
bro where you got whiteboard ?
I mean of course 6 8 10 differ by 2, you just take the 3 4 5 perfect square and multiply everything by 2. You can multiply it by any positive integer n you like to get a perfect square with difference n between the numbers... because if a^2 + b^2 = c^2, surely (an)^2 + (bn)^2 = (cn)^2. Still a cool video though :)
Another related cool property is that an odd prime p is a perfect square if and only if it is of the form p = 4k + 1, and then only in one way (so there is precisely one a and one b such that a^2 + b^2 = p^2)
Also where did you suddenly get 4n+2?
Bit confusing
If b = 2n + 1, then 2b = 4n + 2
@@azulejoa Well that was incredibly stupid of me
Where did he get 3^2 in 3^2=b+1
a=6 => a/2=3
@@PapaFlammy69 Ah ok
It's not really surprising, that the subset of the pythagorean triples you are looking at, doesn't contain a triple for which the side length of a and b are both odd.
There is simply no pythagorean triple for which sides a and b could be both odd:
All pythagorean triples are of the form (u^2-v^2, 2uv, u^2+v^2), u, v in ℕ, u > v.
2uv is never odd.
You can also see this by considering a^2+b^2=c^2 modulus 4.
wo sind die Links zu dem Clip am Anfang und zu dem Mathenewsletter?
Chris Smith's Twitter, alles in der Beschreibung, wie immer
@@PapaFlammy69 merci
n^2 - (n-1)^2 = 2n + 1
By choosing n such that 2n + 1 = a^2,
you can make a pythagorean triple for any odd number greater than 1.
I have been doing this for years before and during my undergrad, I though this was a widely known proof!? The fact a^2=b+c was super helpful during a few modules and proof exercises. Also that Even Pythagorean Triples are not unique in angle/angle/angle measures.
Pythagoras: I m inevitable......
Flammy: Hold my blackboard....
I haven't watched quite yet, Jens, but I can just say that I hope this is a video addressing the metric tensor!!! I'm literally writing my own script on a derivation of its properties to give to the students in GR for my university (RPI). You can derive tensor transformation laws from the more fundamental, tensorial quality of multilinearity. Absolutely brilliant! And, the metric, a tensor field describing the entirety of information regarding length, angle, and curvature {all the intrinsic properties} of a metric space (often just called a manifold) as proven by Gauss's theorema egregium {which uses the connection coefficients [Christoffel Symbols] !!}, is so integral to the nature of reality. I like to call the evaluation of the metric on two vectors in the tangent bundle equating to the first fundamental form: the generalized Pythagorean theorem. Lmao I am nerding out here. Sorry LOL!!! I love you so much, by the way! Thank you for the content and for inspiring me to stay excited and in anticipation!
I've spent the day messing around with cases of this and I'm closing in on a formula for triples of the form (a, b, b+2n) for the natural number n. There's a ton of constraints on it but I'm enjoying the process either way, love ur vids bb
nice!!!
4:20 What?
Love you and your style, dude.
Interesting relationships. a^2-b = c where c=b+1 because for any right triangle, (a^2/(c-b) -b = c.
Where c-b=1, a^2/(c-b) = a^2.
Care to prove (a^2/(c-b))-b=c for any a^2+b^2=c^2? :-)
Here's another good one, For any right triangle, (2b+c-b)*(c-b)=a^2. There. That's not circular. Cool. Proof isn't hard.
Can you make a video explaining formats last theorem proof?
Something I found: if c=b+1, then c-a is two times a square.
Another fact: If the exponent of an odd prime p in c-b is odd, then p divides both a b and c.
It also works if we replace c-b by c-a.
nice!
didn't exactly follow the video... but does the longest side have to be 1 less than the hypotenuse to work?
if not, 6 8 10 doesn't follow this.
36-8= 18 ??
Also, his board is amazing
12:40 you got bamboozled by similar triangles yet again! This example follows trivially from doubling the side lengths of the (3, 4, 5) triangle.
i kinda felt like i was watching a video about a Viking mathematician explaining Pythagoras
Some random manipulations, with a < b < c three sides of a right triangle (c: hypothenuse):
c == b+n --> a^2+b^2 == c^2 == b^2+2bn+n^2
a^2 == (2b+n)n
Case (1) n odd : n=2k
Case (2) n even: n=2k+1
(1): a^2 == 4(b+k)k --> (a/2)^2 == (b+k)k
(2): a^2 == (2(b+k)+1)(2k+1) --> (a^2-1)/2 == bk+(2k+b)(k+1)
----
With b == a+n , get:
c^2 == a^2+b^2 == a^2+(a+n)^2 == 2a(a+n)+n^2 == 2ab+n^2
For any odd integer k greater than 1: k, (k^2-1)/2, (k^2+1)/2 is a triple.
For any even integer k greater than 2: k, (k^2-4)/4, (k^2+4)/4 is a triple.
The first one came from considering when the difference between consecutive square numbers is itself square (I.e. (n+1)^2 - n^2 = k^2).
The second came from considering the difference between squares separated by one values (I.e. (n+2)^2 - n^2 = k^2)
You can carry on like this (so (n+3)^2-n^2 etc), but the conditions you have to place on k to get integer answers get more complicated
Ah, but a (6,8,10) right triangle is really just a disguised (3,4,5) right triangle! You can take any of your odd ones and get even ones by multiplying everything by 2! Because of the properties of similar triangles, scaling all side lengths by the same amount preserves angles.