I stopped the video after 4 minutes. Too complex, while the solution is rather immediate: notice that 8x³−20 is divisible by 4, so that it is a perfect square iff 2x³−5 is a perfect square. Then notice that both expressions have a similar high-order term: 2x³ and 2x⁵, which differ by a factor x², which is a perfect square. So, one can multiply 2x³−5 by x², and the problem is now: 2x⁵−5x² and 2x⁵−2 must both be perfect squares. But for x large enough, their difference is less than twice their square roots, which is about the difference between two consecutive squares. More precisely, 2x⁵−5x² < 2x⁵−2 = k², so that 2x⁵−5x² ≤ (k−1)² = (2x⁵−2) − 2√(2x⁵−2) + 1, which simplifies to x⁴(8x−25) ≤ 9 − 10x² < 0. Hence x ≤ 3. Then check x = 2 and x = 3.
There's no need to plug in numbers to check x=2. The second expression pretty trivially can't be a perfect square for any even x, since it would be guaranteed to have exactly one factor of 2.
Notice: 8x^3 - 20 = 4(2x^3-5). So we want 2x^3-5=a^2 and 2x^5-2=b^2 to be perfect squares at the same time. (ax)^2 = 2x^5-5x^2, so b^2-(ax)^2 = 5x^2-2. I'm not really sure what to do from here... though i know 5x^2-2 must be the difference of two squares.
@@KiLLJoYUA-cam 5x2 - 2 = c2 - b2 Now 5x2 - 2 has to be positive if x > 0 and x is a natural number. So, b2 - c2 > 0 So, b2 > c2, So b > c, So b - 1 >= c, as b and c are natural numbers.
If 8x^3 - 20 is a perfect square, then so is 2x^3 - 5, and so is 2x^5 - 5x^2. If 2x^5 - 2 = n^2 for some n, then 2x^5 - 5x^2 is smaller, so 2x^5 - 5x^2
8x^3-20=n^2 can be simplified into 2x^3-5=k^2 since n has to be even. Then we multiply it by x^2 and deduct from it the second equation 2x^5-2=m^2. We get 5x^2-2=m^2-(xk)^2=(m+xk)(m-xk) Since right hand side factors must divide left hand side, we get 5x^2-2>=m+xk Square both sides: (5x^2-2)^2>=(m+km)^2 > m^2+(x^2) *(k^2) = 4x^5-5x^2-2 This becomes 4x^5-25x^4+15x^2-6 < 0 Then using similar approach used by Michael with derivatives, we get that this polynomial is always positive for x>=6 We then verify cases 1,2,3,4,5. Only solution is 3.
“There are no natural numbers between consecutive natural numbers, that’s kinda well known” - Genius Michael Penn. Jk Michael your videos are great! Thanks for making them
consider a=2x, then the first equation is an elliptic curve m^2=a^3-20 (a Mordell’s equation to be more specific), then it would be much easier to find answers with elliptic curve techniques.
Could be done in a slightly different way. The first thing you have to do is to notice from the second equation that x is an odd number. 2(x^5-1)=n^2=>(x^5-1) must be divisible by 2=>x^5 is odd=>x is odd Then we can substitute x=2k+1 and once again using the second equation get the following: 2(x^5-1)=2(x-1)(x^4+x^3+x^2+x+1) Let’s just replace x^4+x^3+x^2+x+1 with f(x) and do our substitution. 4k*f(x)=n^2 and we face two possibilities: either k is a perfect square or it is not. if it is not a perfect square, then k must divide f(x) or otherwise f(x)==0 (mod k) or using our substitution for x: 5==0 (mod k) this is only possible if k=1 or k=5 (yes, I know that 1 is a perfect square but we can check this case here too). It’s easy to check via the first equation that k=5 is not a solution and then to check that k=1 and x=3 is a solution. so all we have to check is what happens if k is a perfect square (and not equal to 1 ‘cause we checked it before) we can also notice that the next perfect square after 1 is… 4 so x>=9. there I used an inequality for the first equation limiting 2(2k^2+1)^3-5 between (4k^3+3k)^2 and (4k^3+3k+1)^2 and it holds for all k>=4. This way you can get an answer faster and the calculations are a bit easier.
This is a Floor Function problem from the Indian National Maths Olympiad 2014 problem 2 (since you love floor functions so much) The problem states : Let n be a natural number , prove that floor(n/1) + floor(n/2) + floor(n/3) + ... + floor(n/n-1) + floor(n/n) + floor(sqrt(n)) is even , below is the link to official pdf of the problem olympiads.hbcse.tifr.res.in/olympiads/wp-content/uploads/2016/09/inmo-2014.pdf Thanks!
This isn't that bad. It comes down to recognizing that s(n) := floor(n/1) + floor(n/2) + ... + floor(n/n) = d(1) + d(2) + ... + d(n), where d(k) is the number of divisors of k. You can show this by induction. Thus, s(n) changes parity only when d(n) is odd; but this is only the case when n is a perfect square, in which case floor(sqrt(n)) also increases by 1 and so the parity of s(n) + floor(sqrt(n)) never changes. It starts even, so it's always even.
I got lost in Angel's sea of symbols, so I'm probably reiterating what they said. But the key for the induction is calculating s(n+1)-s(n). As before, I define s(n) := floor(n/1) + floor(n/2) + ... + floor(n/n). Then, s(n+1) - s(n) = [floor((n+1)/1) - floor(n/1)] + [floor((n+1)/2) - floor(n/2)] + ... + [floor((n+1)/n) - floor(n/n)] + floor((n+1)/(n+1)). Each difference [floor((n+1)/k) - floor(n/k)] is equal to either 0 or 1, and the difference is equal to 1 precisely when k divides n+1. So we have that s(n+1) - s(n) is equal to the number of divisors of n+1, which I referred to as d(n+1). A simple induction argument from here shows that s(n) = d(1) + d(2) + ... + d(n). (You might refer to this as solving the recursion, though recursion-solving is really just induction.) If we define f(n) := s(n) + floor(sqrt(n)), the quantity we are trying to show is even, then we can establish that f(n) is always even via another induction, by demonstrating that f(1) is even and that f(n+1) - f(n) is always even. f(1) = s(1) + floor(sqrt(1)) = d(1) + 1 = 2, which is even. And f(n+1) - f(n) = [s(n+1) - s(n)] + [floor(sqrt(n+1)) - floor(sqrt(n))] = d(n+1) + [floor(sqrt(n+1)) - floor(sqrt(n))]. If n+1 is a perfect square, then d(n+1) is odd and [floor(sqrt(n+1)) - floor(sqrt(n))] is 1, so their sum is even. If n+1 is not a perfect square, then d(n+1) is even and [floor(sqrt(n+1)) - floor(sqrt(n))] is 0, so their sum is again even. In either case, f(n+1) - f(n) is even. As such, f(n) is even for every positive integer n.
Nice. I have a soln but probably there's a shorter more elegant one? firstly if 1/a + 1/b = 1/c then algebra gives (a+b)c=ab Next, If gcd(a,b,c)=1 this does not mean a,b,c are pairwise coprime but it does mean they can be expressed as a=xyk,b=yzl,c=xzm where x,y,z,k,l,m are pairwise coprime. plugging into formula you get (x^2)yzkm+xy(z^2)lm=x(y^2)zkl dividing by xyz you get xkm+zlm=yzkl i.e. m(xk+zl)=ykl but since m is coprime with all the factors on the RHS m must be 1, so xk+zl=ykl k is coprime with zl but not with xk and ykl so k must equal 1 and x+zl=yl l is coprime with x but not with zl and yl so l=1 and x+z=y so have a=xyk=x(x+z)(1), b=yzl=(x+z)(z)(1) c=xzm=x(z)(1) and a+b=x(x+z)+ z(x+z)=(x+z)^2=y^2
@@jamescollis7650 great there is a solution more elegant : Suppose gcd (a,b)=d So a=dx , b=dy such that gcd(x,y)=1 1/a +1/b =1/c So (a+b)c=ab So ac+bc=ab Replace a by (dx )and b by (dy) So dxc + dyc = (d^2) xy So xc + yc = dxy Let divide by x So c+ yc/x = dy Since that dy is integer and y/x is not divisible (gcd (x,y)=1) so x must divide c By the same proof we prove that y must divide c So c =kxy gcd(x,y)=1 So we replace c by kxy in the equation number (2) K(x^2)y +k (y^2)x =dxy So xy(kx +ky )= dxy So kx +ky =d So k (x+y)=d So x+y=d/k And gcd (d,k)=1 So d/k is not divisible And x+y is integer so d/k must be interger so the only case is k=1 So x+y=d We have a=dx ,b=dy a+b =dx+dy=d (x+y)=d × d= d^2
Nice problem. Have I missed something in this more direct reasoning? From the second equation n is even, and expanding out n=2n' for some n' shows that x is odd. Let x=1+2x' for some x'. Expand the second equation out in terms of x' and n'. Now it seems x' is a factor of n' so write n'=x'n'' for some n''. Expand again and note that x' is a factor of 5, i.e. 1 or 5. Then x=3 or x=11 and only 3 satisfies the first equation.
I do not see how x' is a factor of n'. It looks like x' is a factor of (n')^2 but to conclude that x' is a factor of n', we need an extra fact like x' being prime.
@@numbers93 Yes, I think you're right. n' must share at least half of the occurrences of each prime factor of x', but not necessarily all of them, and that's not enough to support the rest of my argument.
Why did i thought that Since 8x³-20 is perfect square say a² then 2x³-5 should also be perfect square b² such that b=a/2 now 2x^5 -2 =c² Subtracting both 2x^5 - 2x³ +3 =c²-b² c is even cause c² is even while b is odd. c²-b² will be odd c²-b²-3 is even Hence x^5-x³ can be any no. And thats the end lol
The only solution is (4,18,40) I'll try to remember to write the proof some time later. Edit: I just realized that the question says real and not necessarily natural numbers.
this one is pretty tough, assuming you mean 3^m = 7 n^2 - 1. Here's what I cooked up: At the very least, we need 3^m - 1 to be divisible by 7. Using modular arithmetic, we can tell that the equation can only possibly hold when m = 3 + 6k, for some nonnegative integer k. And for each such k, there can be at most 1 possible solution for n, since we can solve n = \sqrt((3^m - 1)/7). So we are left to check for when (3^(3 + 6k) - 1)/7 is a perfect square, which I do not have the tools to tackle completely. When k = 0, we get m = 3, n = 2. There does not seem to be a solution for the next few k values, k=2,3,4,... I tried to work through it for general k and obtained the following result. When k > 0, we must satisfy: 2^3 * 3^3 * 13 * (1 + 3^6 + ... + 3^(6(k-1))) = (n-2)(n+2). That is to say, the factors on the left side have to fit nicely between the factorizations of n-2 and n+2 if we are to have n as an integer. It is especially problematic that the geometric series 1 + 3^6 + ... + 3^(6(k-1)) factors into a cyclotomic polynomials evaluated at 3^6, and though a cyclotomic polynomial is irreducible, its evaluation is not necessarily prime. For example, if k =2, then we have the polynomial 1+x^(k-1) = 1 + x is an irreducible polynomial, but when evaluated at x = 3^6, we get a composite integer 1 + 3^6 = 730. Edit: After putting further thought into it, I realized that (m,n) = (3,2) is the only solution. We can use number theory and a contradiction argument to show that the equation I arrived on can never be satisfied for any k > 0.
Once established that m = 6k + 3, as showed by Dan Q, one can proceed as follows: (1) Rewrite the equation 3^m = 7n^2 - 1 as x^3 + 1 = 7n^2, where x = 3^{2k+1}; (2) Factor x^3 + 1 as (x + 1)(x^2 - x + 1); (3) Note that x^2 - x + 1 = (x + 1)(x - 2) + 3; therefore, if p divides x + 1, then p doesn't divide x^2 - x + 1 (unless p = 1 or p = 3, but we can eliminate the latter case, since 3 doesn't divide x + 1 = 3^{2k+1} + 1), and vice-versa; (4) Since x + 1 and x^2 - x + 1 doesn't have prime factors in common, the equation (x + 1)(x^2 - x + 1) = 7n^2 can only be solved over the integers if one of the following situations is satisfied: (4.1) x + 1 = 7a^2 and x^2 - x + 1 = b^2, or (4.2) x + 1 = a^2 and x^2 - x + 1 = 7b^2, where a and b are coprime; besides, 7 cannot be a factor of b in the first case, or a factor of a in the second. (5) Solving the quadratic equation in case (4.1), we obtain x = [1 +- sqrt(4b^2 - 3)]/2; in order that x be an integer, we must have 4b^2 - 3 = c^2, where c is a positive integer. Rearranging the terms we obtain 4b^2 - c^2 = 3, or (2b - c)(2b + c) = 3, which implies 2b - c = 1 and 2b + c = 3, or b = c = 1. It follows that x = 0 or x = 1, but none of these solutions is acceptable, since x = 3^{2k+1}. (6) In case (4.2), let's first solve the equation x + 1 = a^2: a^2 - 1 = x = 3^{2k+1} => (a - 1)(a + 1) = 3^{2k+1} => a - 1 = 3^u and a + 1 = 3^v, where u + v = 2k + 1 and v > u >= 0. It's not difficult to show that the only integer solution to those equations is u = 0, v = 1 (so that k = 0) and a = 2, which implies x = 3. Inserting this value in the equation x^2 - x + 1 = 7b^2, we obtain 7 = 7b^2, so that b = 1. Finally, we obtain m = 6k + 3 = 3 and n = ab = 2 as the only positive integer solution to 3^m = 7n^2 - 1.
Hi Ted, I will break this down for you. So first make yourself aware, that the lower order terms become diminishingly small compared to higher order terms, when x grows beyond certain numbers. (Let's imagine x=10 just for the sake of it.) So Michael is now looking for some perfect square that is near the polynomial. For that he is taking the perfect square, that gives us the first two terms of the polynomial. To get it, he takes the square root of 16x^8, which is 4x^4 and now looks for the second part of the binomial that will give him -40x^5, when multiplying out. So he got -5x. (4x^4 - 5x)^2 = 16x^8 - 40x^5 + 25x^2 , so he had to add a -25x^2 in the later part so he does not change the overall sum. He then also put all 3 latter terms into brackets with negative sign in the very same line. So now we know, that n²m² is very near (4x^4-5x)^2 , but not the same, as only a smaller order term is subtracted. When going on, he now looks at the previous perfect square (4x^4-5x-1)^2 , which for same reasons must also be somewhat near n²m², as only a smaller order term is added. The exercise with f(x) and g(x) is to show that indeed (4x^4-5x)^2 > n²m² > (4x^4-5x-1)^2.
Here is a fancy pants solution. The equation m^2=8x^3-20=(2x)^3-20 defines a Mordell-type elliptic curve. There are algorithms that quickly give the integer solutions to these. Applying one to this equation, we get that the only integer solution is (2x,m)=(6,14). And when x=3, it turns out incidentally that 2x^5-2 is a square.
I stopped the video after 4 minutes. Too complex, while the solution is rather immediate: notice that 8x³−20 is divisible by 4, so that it is a perfect square iff 2x³−5 is a perfect square. Then notice that both expressions have a similar high-order term: 2x³ and 2x⁵, which differ by a factor x², which is a perfect square. So, one can multiply 2x³−5 by x², and the problem is now: 2x⁵−5x² and 2x⁵−2 must both be perfect squares. But for x large enough, their difference is less than twice their square roots, which is about the difference between two consecutive squares. More precisely, 2x⁵−5x² < 2x⁵−2 = k², so that 2x⁵−5x² ≤ (k−1)² = (2x⁵−2) − 2√(2x⁵−2) + 1, which simplifies to x⁴(8x−25) ≤ 9 − 10x² < 0. Hence x ≤ 3. Then check x = 2 and x = 3.
Why is the "difference less than twice their square roots"? Could you please explain?
solid work!
Much more straightforward than Michael's solution, I like it
There's no need to plug in numbers to check x=2. The second expression pretty trivially can't be a perfect square for any even x, since it would be guaranteed to have exactly one factor of 2.
This is too good... you just demolished this problem only to do an arithmetic mistake in the last part...8-20 = -16.
Yeah the second I saw that I was confused lol.
3:54 -39➡️+39
I liked the part where 8-20=-16.
UA-cam should have recommended me this channel sooner!!
When you watch a few similar videos you get the idea and it is supeeeer helpful
Shouldn't G have +39 instead of -39?
I know right?
Notice: 8x^3 - 20 = 4(2x^3-5). So we want 2x^3-5=a^2 and 2x^5-2=b^2 to be perfect squares at the same time. (ax)^2 = 2x^5-5x^2, so b^2-(ax)^2 = 5x^2-2. I'm not really sure what to do from here... though i know 5x^2-2 must be the difference of two squares.
Here is another approach: start with
2x3-5 = a2
2x5-2 = b2
mpy top by x2:
2x5-5x2 = c2
then subtract from second line
5x2-2 = b2-c2
use
c
very good
Can you help me solve this problem?
find all positive integer m，n，s.t. :
3＾m=7n²-1
What’s the c = b-1 part?
@@KiLLJoYUA-cam well 5x2-2 is positive so b>c
@@KiLLJoYUA-cam
5x2 - 2 = c2 - b2
Now 5x2 - 2 has to be positive if x > 0 and x is a natural number.
So, b2 - c2 > 0
So, b2 > c2,
So b > c,
So b - 1 >= c, as b and c are natural numbers.
Apparently UA-cam doesn’t want me to post the timestamp today 🤷♂️ My comment is removed again and again...
Probably an attempt to fight the spam comments that always include a single timestamp
@@z4zuse Yeah, that’s probably it.
@@goodplacetostop2973 oh sedlyf
Maybe thats a good place to stop.
might need to rewatch this, hehe, can't understand the inequalities part
If 8x^3 - 20 is a perfect square, then so is 2x^3 - 5, and so is 2x^5 - 5x^2. If 2x^5 - 2 = n^2 for some n, then 2x^5 - 5x^2 is smaller, so 2x^5 - 5x^2
Pretty creative solution
8x^3-20=n^2 can be simplified into 2x^3-5=k^2 since n has to be even. Then we multiply it by x^2 and deduct from it the second equation 2x^5-2=m^2.
We get 5x^2-2=m^2-(xk)^2=(m+xk)(m-xk)
Since right hand side factors must divide left hand side, we get 5x^2-2>=m+xk
Square both sides: (5x^2-2)^2>=(m+km)^2 > m^2+(x^2) *(k^2) = 4x^5-5x^2-2
This becomes 4x^5-25x^4+15x^2-6 < 0
Then using similar approach used by Michael with derivatives, we get that this polynomial is always positive for x>=6
We then verify cases 1,2,3,4,5. Only solution is 3.
14:02
Oh now that works... Weird.
@@goodplacetostop2973 yay it works, your efforts for posting time stamp worked.
“There are no natural numbers between consecutive natural numbers, that’s kinda well known” - Genius Michael Penn.
Jk Michael your videos are great! Thanks for making them
consider a=2x, then the first equation is an elliptic curve m^2=a^3-20 (a Mordell’s equation to be more specific), then it would be much easier to find answers with elliptic curve techniques.
I would never have found that method. The key seems to be choosing that 25. Is that guesswork? Or is there a process?
Could be done in a slightly different way.
The first thing you have to do is to notice from the second equation that x is an odd number.
2(x^5-1)=n^2=>(x^5-1) must be divisible by 2=>x^5 is odd=>x is odd
Then we can substitute x=2k+1 and once again using the second equation get the following:
2(x^5-1)=2(x-1)(x^4+x^3+x^2+x+1)
Let’s just replace x^4+x^3+x^2+x+1 with f(x) and do our substitution.
4k*f(x)=n^2
and we face two possibilities: either k is a perfect square or it is not.
if it is not a perfect square, then k must divide f(x) or otherwise f(x)==0 (mod k) or using our substitution for x:
5==0 (mod k)
this is only possible if k=1 or k=5 (yes, I know that 1 is a perfect square but we can check this case here too). It’s easy to check via the first equation that k=5 is not a solution and then to check that k=1 and x=3 is a solution.
so all we have to check is what happens if k is a perfect square (and not equal to 1 ‘cause we checked it before)
we can also notice that the next perfect square after 1 is… 4 so x>=9.
there I used an inequality for the first equation limiting 2(2k^2+1)^3-5 between (4k^3+3k)^2 and (4k^3+3k+1)^2 and it holds for all k>=4.
This way you can get an answer faster and the calculations are a bit easier.
Your videos are amazing. The problems are very interesting and you are a great mathematician. Awsome solutions...
4:23 should be +39 instead of -39 right?
Mr penn you are doing a great job for the world comunity improvement, you should be awarded in some way many thanks
Nice problem. Thanks.
you are doing great job, I think everyone will agree. Go on, love from India❤👌
Just watch this Math channel..... very impressive channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
This is a Floor Function problem from the Indian National Maths Olympiad 2014 problem 2 (since you love floor functions so much)
The problem states : Let n be a natural number , prove that floor(n/1) + floor(n/2) + floor(n/3) + ... + floor(n/n-1) + floor(n/n) + floor(sqrt(n)) is even , below is the link to official pdf of the problem
olympiads.hbcse.tifr.res.in/olympiads/wp-content/uploads/2016/09/inmo-2014.pdf
Thanks!
This isn't that bad. It comes down to recognizing that s(n) := floor(n/1) + floor(n/2) + ... + floor(n/n) = d(1) + d(2) + ... + d(n), where d(k) is the number of divisors of k. You can show this by induction. Thus, s(n) changes parity only when d(n) is odd; but this is only the case when n is a perfect square, in which case floor(sqrt(n)) also increases by 1 and so the parity of s(n) + floor(sqrt(n)) never changes. It starts even, so it's always even.
@@CauchyIntegralFormula cam u elaborate how is it the divisor of k ?
Just watch this Math channel..... very impressive channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
I got lost in Angel's sea of symbols, so I'm probably reiterating what they said. But the key for the induction is calculating s(n+1)-s(n).
As before, I define s(n) := floor(n/1) + floor(n/2) + ... + floor(n/n). Then, s(n+1) - s(n) = [floor((n+1)/1) - floor(n/1)] + [floor((n+1)/2) - floor(n/2)] + ... + [floor((n+1)/n) - floor(n/n)] + floor((n+1)/(n+1)). Each difference [floor((n+1)/k) - floor(n/k)] is equal to either 0 or 1, and the difference is equal to 1 precisely when k divides n+1. So we have that s(n+1) - s(n) is equal to the number of divisors of n+1, which I referred to as d(n+1). A simple induction argument from here shows that s(n) = d(1) + d(2) + ... + d(n). (You might refer to this as solving the recursion, though recursion-solving is really just induction.)
If we define f(n) := s(n) + floor(sqrt(n)), the quantity we are trying to show is even, then we can establish that f(n) is always even via another induction, by demonstrating that f(1) is even and that f(n+1) - f(n) is always even. f(1) = s(1) + floor(sqrt(1)) = d(1) + 1 = 2, which is even. And f(n+1) - f(n) = [s(n+1) - s(n)] + [floor(sqrt(n+1)) - floor(sqrt(n))] = d(n+1) + [floor(sqrt(n+1)) - floor(sqrt(n))]. If n+1 is a perfect square, then d(n+1) is odd and [floor(sqrt(n+1)) - floor(sqrt(n))] is 1, so their sum is even. If n+1 is not a perfect square, then d(n+1) is even and [floor(sqrt(n+1)) - floor(sqrt(n))] is 0, so their sum is again even. In either case, f(n+1) - f(n) is even. As such, f(n) is even for every positive integer n.
Thank you!
We could also observe that if 8x^3-20 is a perfect square then so is 2x^5-5x^2, and since cx^2
Just watch this Math channel..... very impressive channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
Is this a case of "law of small numbers"? Obviously expressions with x^5 shouldn't hit perfect squares often
Look at this problem : (a,b,c)are positive integer
If :
Gcd (a,b,c)=1 and 1/a +1/b =1/c
Prove that a+b is a perfect square
Nice. I have a soln but probably there's a shorter more elegant one?
firstly if 1/a + 1/b = 1/c then algebra gives (a+b)c=ab
Next, If gcd(a,b,c)=1 this does not mean a,b,c are pairwise coprime but it does mean they can be expressed as a=xyk,b=yzl,c=xzm where x,y,z,k,l,m are pairwise coprime.
plugging into formula you get
(x^2)yzkm+xy(z^2)lm=x(y^2)zkl
dividing by xyz you get
xkm+zlm=yzkl i.e. m(xk+zl)=ykl
but since m is coprime with all the factors on the RHS m must be 1, so
xk+zl=ykl
k is coprime with zl but not with xk and ykl so k must equal 1 and
x+zl=yl
l is coprime with x but not with zl and yl so l=1 and
x+z=y
so have a=xyk=x(x+z)(1), b=yzl=(x+z)(z)(1)
c=xzm=x(z)(1)
and a+b=x(x+z)+ z(x+z)=(x+z)^2=y^2
@@jamescollis7650 great there is a solution more elegant :
Suppose gcd (a,b)=d
So a=dx , b=dy such that gcd(x,y)=1
1/a +1/b =1/c
So (a+b)c=ab
So ac+bc=ab
Replace a by (dx )and b by (dy)
So dxc + dyc = (d^2) xy
So xc + yc = dxy
Let divide by x
So c+ yc/x = dy
Since that dy is integer and y/x is not divisible (gcd (x,y)=1) so x must divide c
By the same proof we prove that y must divide c
So c =kxy gcd(x,y)=1
So we replace c by kxy in the equation number (2)
K(x^2)y +k (y^2)x =dxy
So xy(kx +ky )= dxy
So kx +ky =d
So k (x+y)=d
So x+y=d/k
And gcd (d,k)=1
So d/k is not divisible
And x+y is integer so d/k must be interger so the only case is k=1
So x+y=d
We have a=dx ,b=dy
a+b =dx+dy=d (x+y)=d × d= d^2
@@hamzasrayhe9431 Agree that is more elegant 👍Doesn't require quite so many letters!
how for us to come up with the claim x >=4?
I mean how to think of the number 4 in the first hand.
TY
Nice problem. Have I missed something in this more direct reasoning? From the second equation n is even, and expanding out n=2n' for some n' shows that x is odd. Let x=1+2x' for some x'. Expand the second equation out in terms of x' and n'. Now it seems x' is a factor of n' so write n'=x'n'' for some n''. Expand again and note that x' is a factor of 5, i.e. 1 or 5. Then x=3 or x=11 and only 3 satisfies the first equation.
Just watch this Math channel..... very impressive channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
I do not see how x' is a factor of n'. It looks like x' is a factor of (n')^2 but to conclude that x' is a factor of n', we need an extra fact like x' being prime.
@@numbers93 Yes, I think you're right. n' must share at least half of the occurrences of each prime factor of x', but not necessarily all of them, and that's not enough to support the rest of my argument.
You made me love mathematics 😋
Just watch this channel... Very impressive Math channel... ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
That's impressive, I thought this really was more one for hard core math people
@@romajimamulo yeah in the beginning he made video for AL students but now more complex..
Thanks. It means a lot for me.
It’s kind of weird that 6*18 = 108… and 56*18=1008 and 556*18 = 10008. The pattern seems to continue
As 8 and 20 are divisible by 4, 8x^3-20 is a perfect square if and only if 2x^3-5 is one too. The calculations could be quite easier.
More number theory!
I always associate x with the reals, really through me in the thumbnail.
What will happen, if we put 7x^3 instead of 8x^3 in a first equation? This solution doesn't works in this case, but using LTE we can easily bash it
Just watch this Math channel..... very impressive channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
At __ : __ coming soon...
Why did i thought that
Since 8x³-20 is perfect square say a² then 2x³-5 should also be perfect square b² such that b=a/2
now 2x^5 -2 =c²
Subtracting both
2x^5 - 2x³ +3 =c²-b²
c is even cause c² is even while b is odd.
c²-b² will be odd
c²-b²-3 is even
Hence x^5-x³ can be any no.
And thats the end lol
Please include this problem in one of your videos
Let x,y,z be reals in the interval [4,40] such that :
x+y+z=62
xyz=2880
find all triples (x,y,z).
The only solution is (4,18,40) I'll try to remember to write the proof some time later.
Edit: I just realized that the question says real and not necessarily natural numbers.
@@Ooopsi ok
@@imobusters4049 Michael Penn made a video about this one a moment ago
Some stuffs come from nowhere. It looks like "shut up, it's magic!".
Agree completely
michael. could you help me solve this problem ? m,n is positive integer. S.T : 3＾m=7b²-1
this one is pretty tough, assuming you mean 3^m = 7 n^2 - 1. Here's what I cooked up:
At the very least, we need 3^m - 1 to be divisible by 7. Using modular arithmetic, we can tell that the equation can only possibly hold when m = 3 + 6k, for some nonnegative integer k. And for each such k, there can be at most 1 possible solution for n, since we can solve n = \sqrt((3^m - 1)/7). So we are left to check for when (3^(3 + 6k) - 1)/7 is a perfect square, which I do not have the tools to tackle completely.
When k = 0, we get m = 3, n = 2. There does not seem to be a solution for the next few k values, k=2,3,4,... I tried to work through it for general k and obtained the following result. When k > 0, we must satisfy:
2^3 * 3^3 * 13 * (1 + 3^6 + ... + 3^(6(k-1))) = (n-2)(n+2).
That is to say, the factors on the left side have to fit nicely between the factorizations of n-2 and n+2 if we are to have n as an integer.
It is especially problematic that the geometric series 1 + 3^6 + ... + 3^(6(k-1)) factors into a cyclotomic polynomials evaluated at 3^6, and though a cyclotomic polynomial is irreducible, its evaluation is not necessarily prime. For example, if k =2, then we have the polynomial 1+x^(k-1) = 1 + x is an irreducible polynomial, but when evaluated at x = 3^6, we get a composite integer 1 + 3^6 = 730.
Edit: After putting further thought into it, I realized that (m,n) = (3,2) is the only solution. We can use number theory and a contradiction argument to show that the equation I arrived on can never be satisfied for any k > 0.
@@numbers93 Thank you very much. I try my best to understand your answer
Once established that m = 6k + 3, as showed by Dan Q, one can proceed as follows:
(1) Rewrite the equation 3^m = 7n^2 - 1 as x^3 + 1 = 7n^2, where x = 3^{2k+1};
(2) Factor x^3 + 1 as (x + 1)(x^2 - x + 1);
(3) Note that x^2 - x + 1 = (x + 1)(x - 2) + 3; therefore, if p divides x + 1, then p doesn't divide x^2 - x + 1 (unless p = 1 or p = 3, but we can eliminate the latter case, since 3 doesn't divide x + 1 = 3^{2k+1} + 1), and vice-versa;
(4) Since x + 1 and x^2 - x + 1 doesn't have prime factors in common, the equation (x + 1)(x^2 - x + 1) = 7n^2 can only be solved over the integers if one of the following situations is satisfied:
(4.1) x + 1 = 7a^2 and x^2 - x + 1 = b^2, or
(4.2) x + 1 = a^2 and x^2 - x + 1 = 7b^2,
where a and b are coprime; besides, 7 cannot be a factor of b in the first case, or a factor of a in the second.
(5) Solving the quadratic equation in case (4.1), we obtain x = [1 +- sqrt(4b^2 - 3)]/2; in order that x be an integer, we must have 4b^2 - 3 = c^2, where c is a positive integer. Rearranging the terms we obtain 4b^2 - c^2 = 3, or (2b - c)(2b + c) = 3, which implies 2b - c = 1 and 2b + c = 3, or b = c = 1. It follows that x = 0 or x = 1, but none of these solutions is acceptable, since x = 3^{2k+1}.
(6) In case (4.2), let's first solve the equation x + 1 = a^2:
a^2 - 1 = x = 3^{2k+1} => (a - 1)(a + 1) = 3^{2k+1} => a - 1 = 3^u and a + 1 = 3^v, where u + v = 2k + 1 and v > u >= 0. It's not difficult to show that the only integer solution to those equations is u = 0, v = 1 (so that k = 0) and a = 2, which implies x = 3. Inserting this value in the equation x^2 - x + 1 = 7b^2, we obtain 7 = 7b^2, so that b = 1. Finally, we obtain m = 6k + 3 = 3 and n = ab = 2 as the only positive integer solution to 3^m = 7n^2 - 1.
@@ricardocavalcanti3343 thank you very much，i have read your idea of solving the problem. it is perfect
I'm just an ordinary person, so I have no idea whatsoever what happened at around 2:50
Hi Ted, I will break this down for you. So first make yourself aware, that the lower order terms become diminishingly small compared to higher order terms, when x grows beyond certain numbers. (Let's imagine x=10 just for the sake of it.) So Michael is now looking for some perfect square that is near the polynomial. For that he is taking the perfect square, that gives us the first two terms of the polynomial.
To get it, he takes the square root of 16x^8, which is 4x^4 and now looks for the second part of the binomial that will give him -40x^5, when multiplying out. So he got -5x.
(4x^4 - 5x)^2 = 16x^8 - 40x^5 + 25x^2 , so he had to add a -25x^2 in the later part so he does not change the overall sum. He then also put all 3 latter terms into brackets with negative sign in the very same line.
So now we know, that n²m² is very near (4x^4-5x)^2 , but not the same, as only a smaller order term is subtracted.
When going on, he now looks at the previous perfect square (4x^4-5x-1)^2 , which for same reasons must also be somewhat near n²m², as only a smaller order term is added.
The exercise with f(x) and g(x) is to show that indeed (4x^4-5x)^2 > n²m² > (4x^4-5x-1)^2.
@@petersievert6830 what an incredibly kind gesture for a stranger on the internet.
Thank you very much.
Hi Michael ... Do you have a Patreon account?
Just watch this Math channel..... very impressive channel ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
Here is a fancy pants solution. The equation m^2=8x^3-20=(2x)^3-20 defines a Mordell-type elliptic curve. There are algorithms that quickly give the integer solutions to these. Applying one to this equation, we get that the only integer solution is (2x,m)=(6,14). And when x=3, it turns out incidentally that 2x^5-2 is a square.
A perfect square never ends 2 3 7 or 8
8-20 = -12 not -16
15 minutes too late?
What will you do if the inequality will leave you testing for 1,000,000 or more numbers, sir? You just got lucky here.
If you expect any one technique to work for all problems, get ready to be disappointed.
You can use a computer to test those cases.