Lol, it’s exactly what I did, just watch the thumbnail, was like « Ok (1;1) is a solution when x=y » and then... « If you multiply the second equation by 5 we get a polynomial expression that we can factorise by (x-y) and something of degree 2...damn I’ve got to try it » and so I did lol, it works fine.
6:30 instead of doing polynomial division, if you add y - y and split the -2y^2 into -y^2 and -y^2 you get (y^3 - y^2) - (y^2 - y) - (y + 1). You can now see that you can take out a (y-1) from each of those. I got this, but admittedly I did do the polynomial division first then looked for a prettier solution once I knew the end result, but it’s still cool imo: A reminder that factoring non-obvious polynomials like this is in some ways about “creating or finding the symmetry” which I think is a helpful way to look at it for other factoring problems too.
Plz do a collab on a topic of math plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
about the division made in 8:00, I can affirm that in Portugal, my country, we learn it in high school (10th grade), but it is called as Ruffini's Rule
6:32, I called this "Ruffini's rule", a particular version of "Horner's method" ( usually called synthetic division). I remembered I learned this in highschool, and I used to do it without thinking. Then I understood, how division of polynomials work, and never used it again until the point I completely forgot about the algorithm.
On 7:59, Bprp, you said "I'm not sure whether this is a popular method". Yes, in Indonesia, the method (If I am not mistaken: Horner Method) was popular just for a while (maybe until 2013). But since Indonesian math curriculum did not add this in the newest curriculum (which is from 2014), not many know this method.
What I did is that I first took x^3 common from the first equation and y^3 from the 2nd.Now I took (y/x) as some value 'a' and divided the two equations.When we simplify this we get a cubic equation having only the value 'a'. If u solve that equation, we get many more terms like the solutions of x^3 = 1 and y^3 = 1 and some more, which also includes the answers u found.But yours was an elegant method and much more easier. Great videos btw
"Hopefully we can get points for this question" 😅👍 yes you do Steve. And I loveeeeeeee your little pi figure from 3blue1brown? And excellent video of course.
You can also divide equation 1 by equation 2 to reduce it to a cubic in terms of t=x/y however substituting back in to solve for x and y is much messier
(1,1) is really easy to find just by inspection, especially if you factor out an x^2 and y^2 from each equation. Then you can skip straight to finding the other two solutions.
The system is symmetric with respect to the exchange of x and y. Then replacing y by x in either of the equations gives us x³=1. Then it is straightforward to find the three roots of x. The real solution is x=1, and the two complex roots are 120° symmetric with respect to the real root, on the complex plane.
By inspection I noted that x = 1 and y = 1 are solutions. I then factored out the solution for x or y as (x-1) or (y-1) and then solved the system for the other solutions.
Ahhh that's very clever! Finding (1,1) by inspection was not very difficult, but I forgot about factoring out those solutions to reduce each equation to a problem in quadratics. Excellent!
Here's a nice trick : Multiply the bottom equation by 5 and subtract it from the first equation. The resulting equation is equal to zero. Now notice x = y is a root of the equation since the sum of coefficients is zero. Divide out the factor (x-y) using long division. Now solve for the other factors using the quadratic formula! Edit : Don't forget to substitute into either of the first two equations to find the 'initial condition' of the system
Could you elaborate on this please? (x-y)(x^2 + 10xy + 5y^2)=0. Is that what you meant by dividing out the factor (x-y) from the equation? If so I don't see how I can apply the quadratic formula or how it helps me, I would be very grateful for your assistance. On further reflection using your method I get y= cuberoot(1/(-2 +- sqrt(5)) which gives the right answer but not in the same form.
Rearranging gives us a^2-ax+x=0 We will search for integer values of x where the solution a is an integer too. a=(x+-sqrt(x^2-4x))/2 For a to be integer we should have no square roots, thus x^2-4x must be a perfect square. x^2-4x can be written as (x-2)^2-4, which is some perfect square minus 4. Now this number is only a perfect square when it is equal to zero (can someone help me prove this?) x^2-4x=0 gives x=0 and x=4 Quickly checking gives us for x=0 a=0 and for x=4 a=2 I dont think I missed any solutions, I just need to prove that (x-2)^2-4 is only a perfect square when equal to zero, can someone tell me how to do it? Edit - nvm I figured it out, here you go: The question is: When is x(x-4)=n^2 for some integers x and n (n larger than zero) we quickly notice that for this to be the case, our 2 factors must be related to n,one of them must be divisible by n and the other must be a divisor of n , the larger should be divisible by n, so x is divisible by n and x-4 is a divisor of n by the same coefficient: meaning that x=bn for some integer b and n=b(x-4) for the same value of b. substituting n in the first equation: x=b[b(x-4)]=b^2(x-4) x=(b^2)x-4b^2 thus x(b^2-1)=4b^2 x=(4b^2)/(b^2-1) we can factor b^2-1 into (b-1)(b+1) and 4b^2 is (2b)^2 since x=((2b)^2)/(b-1)(b+1) and since x is an integer then (2b)^2 is divisible by both b-1 and b+1 using the logic we used earlier b+1=c(2b) and b-1=(2b)/c b+1=c^2(b-1) b(c^2-1)=c^2+1 b=(c^2+1)/(c^2-1) b=((c^2-1)+2)/(c^2-1) b=1+(2/(c^2-1)) since b is an integer and 1 is an integer, 2/(c^2-1) should also be an integer. c^2-1 has a minimum value of -1, which gives b=-1, and b=-1 gives no x solutions. For c=1 b is undefined (1+2/0) , for c greater than or equal to 2, c^2-1 is greater than 2, denominator greater than numerator aka 2/(c^2-1) wouldn't be an integer. Thus b can't be an integer and by contradiction there are no solutions (other than the case where n=0, which gives x=0 or x=4) One more thing, x=0 and a=0 is not a valid solution, since 0/0 is undefined. Thus the only solution we have is x=4 a=2 Edit 2 - I just noticed that c isn't necessarily an integer, but c^2 is. Solution isn't affected much though, no new x values appear. Edit 3 - Apparently b^2 and b do not have to be integers, but c^2 has to, so we can just substitute integer values of c^2 and check if we get a valid x solution. If anyone has a more efficient way of doing this please tell me.
Nevermind I found a much easier solution. Since both x and a are integers, x-a is an integer, thus x must be divisible by a. Let x=an for some integer n, we have: an/a=an-a n=an-a n=a(n-1) a=n/n-1 Now we need a to be an integer, so we can try some n values. For n=0 a=0 (rejected) For n=1 a=1/0 (rejected) For n=2 a=2, which gives x=4 (accepted solution) For n above 2, n-1 becomes greater than 1, n and n-1 would be relatively prime, and thus we wouldn't be able to divide n by n-1 and get an integer. For negative values of n, n=-1 a=1/2 n=-2 a=2/3 Etc Thus the only solution is the pair x=4 and a=2
This night I weirdly dreamt about watching a long bprp video and forty minutes in we start hearing violin and he's kinda mad since it's his children and they are not practicing the piece they're supposed to. Given that Steve seemed the same as he is now, his children were likely very young while playing at professional level. Weird dream, but fun dream.
About that dividing polynomials method you used in 8:00 what i can confirm is in Poland where i live about 10 years ago while i was in high school we had it but it was class with advanced math profile, classes with basic math didn't have it and last time i checked like 2-3 years ago advanced classes didn't have that method anymore
Are x and y real numbers?(I couldn't find the conditions). If complex numbers are also possible, x=y=(-1+sqrt(3)*i) / 2 and x=y=(-1-sqrt(3)*i) / 2 can be solutions as well.
Красиво. Особенно разложение на множители. Разделил оба на xy^2, замена x/y=u, доумножил на -5, сложил, получил кубическое уравнение u^3+4u-5=0. Дальше, только разглядев u=1, смог разложить на множители, без подсказки видео не додумался. Здоровское решение, спасибо.
since the given equations are homogenous in nature put y=kx, later eliminate x^3 find values of k, we will get one solution as k=1 and another quadratic equation of k, let the roots be k1,k2,k3 we will get y1,y2,y3, and substitute in the given equations, we will get x1,x2,x3 now again put in y1=k1*x1, similarly y2=k2*x2, and y3 also
What you got are full real solutions of the problem, but if we consider the complex number field, I think that there can be three cases in 4:21 As you know in complex field, (x+3y)^3 = 4^3 has three cases. x+3y = 4 or x+3y = 4e^{2pi i/3} or x+3y = 4e^{4pi i / 3} I don't know full statement of that problem in original olympiad contest. But if we want to find full 'complex' solutions, it would be troublesome!
x³ + 9x²y = 10 y³ + xy² = 2 First blush of my thought process: Hmmm..the LHS of the 1st equation looks like the start of the expansion of (x + 3y)³ - what's the rest? Let's see, x³ + 3·3x²y + 3·9xy² + 1·27y³ = x³ + 9x²y + 27xy² + 27y³ = x³ + 9x²y + 27(xy² + y³) And son-of-a-gun! There's the LHS of the 2nd equation! So (x + 3y)³ = 10 + 27·2 = 64 x + 3y = 4 [or 4 times either of the two complex roots of unity] Hey! How about x = y = 1? YES!! That works! That could have been guessed without going through any of that rigamarole above. What about other possible solutions, even without using the complex roots we set aside earlier? y = ⅓(4-x) 10 = x³ + 3x²(4-x) = -2x³ + 12x² x³ - 6x² + 5 = 0
@@midorimashintaro2613 Some are available on my Mac keyboard using Option and Option-Shift (in Windows, "alt" & "alt-Shift"); some I've collected from other posts on Yahoo Answers & YT, by Copy-Paste into a text file, then from there to YT comments. In Windows, I understand there are ways to get these Unicode characters using the keyboard... Fred
It looks give more questions than answers. How to calculate the number of roots/solutions for this equation (it looks like having total 5 roots in WFA)? Why can you divide the equations (1) and (2) with (x-1)*(y-1) and then solve rest of the roots?
Since both equations are homogenous. Let y = mx for some real number m and substitute that in both equations. This will eliminate the y variable. Now factor x^3 from both equations and divide the equations . Now youll get the cubic polynomial in m . Solve for m and back substitute to get values of x & y for corresponding values of m.
Blackpenredpen did u can solve the question in by like 2 seconds by using cross method multiplication. (10*1-9*2)/1*1-9*1= - 8/-8 = 1 thus the first row =1 therefore x and y equal 1(the row only works in this case because of x and y having no coefficient usually it is only x and u sub in to get y)
Funny thing is, I "substituted right away" x = 1, y = 1. Another attack is to divide the first equation by the second to get (t^3 + 9t^2)/(1+t) = 5 where t = x/y. So t^3 + 9t^2 - t - 5 = 0 where we know t-1 divides the LHS since the solution x=1, y=1 gives t = 1 as a root. The other two solutions (after synthetic division, quadratic formula) are t = -5 + 2\sqrt{5}, t = -5 - 2\sqrt{5}. It's not too hard from there. Also one shouldn't worry about real versus complex: by homogeneity of the left-hand sides, if (x, y) is a solution, then so is (cx, cy) where c is a cube root of unity.
8:00 "I'm not sure this is the popular method" Popular? In middle & high school every student hated Ruffini's rule ahah. But yes, it's a good quick method I think
Solving the second equation for x, and substituting into the first equation is not very difficult. You get a cubic in y^3, and y^3=1 is one of the solutions. Divide it out, and you get a quadratic in y^3. If I didn't make a mistake, y^3 = 2 +- sqrt(5) is the other pair of solutions. x = (2 - y^3)/y^2 -- Now getting all pairs (x,y) might be a bit more grungy.
That's a very cool way to solve this system. I multiplied the bottom one by 5 and then subtracted it from the top one to get a homogeneous equation, however it gets kinda messy after solving for x/y. It's not that hard, but that's a definitely more arithmetic involved approach. It would be nice to see more geometry on your channel as well, btw
@@rahimeozsoy4244 well, a function is called homogeneous if it has this property: f(ax)=(a^n)f(x), and n is called a degree of homogeneity. It's true for homogeneous differential equations as well but that's not the only case. For any homogeneous equation of two variables it is true that dividing equation by y^n (or x^n, it's symmetric) you will get a polynomial of degree n with respect to x/y (or y/x). Any monomial is homogeneous (it's rather obvious) and every polinomial that consists of monomials of same degree is homogeneous too (for that it should not include any nonzero constants). So in a sense, calling something homogeneous means that this function/equation is symmetrical (loosely speaking) upon change of x and y and proportional too their multiples. I like to think about it as it means that some sort of equivalency in geometry (homothety). It's equal to proportion. So having everything in terms of x/y (y/x) makes sense, you are concerned only with ratios, not actual values, and after figuring out ratios you can easily (in a mathematical sense) figure out values. That's a brief overview, but I recommend to read a little bit about homogeneous functions on your own, it's not a particularly difficult topic. Oh, and there's no need to apologise for any mistakes, English isn't my native language as well, as suggested by my name :), and even if it was it's not a big deal.
I would rather multiply the second equation with 5 and then eliminate the constants by subtracting equation. Then divide the formula by y³ and you get a cubic equation in q=x/y. Then you solve for q, and plugin x=qy in any of the given equation and find x and y. I think this is better because if you get uglier constants it will work
There are 9 solutions x'=cos(2π/3), y'=isin(2π/3) and x"=cos(4π/3), y"=isin(4π/3) are olso other solutions for (1,1). for every solutions (including imaginary numbers) just rotate each solution by 2π/3 and 4π/3 {it means multiply e^(2πi/3) and e^(4πi/3)}
Like the new mic?
I love it!
Black pen red pen Yeah!!!
blackpenredpen yeee
Awww It looks like the pi creatures from 3 blue 1 brown
The new mic is awesome😍...3blue1brown will be happy😄...
“Let me put y iiiiin blue.”
“No why would I do that, I’m blackpenredpen.”
Nole Cuber Lol
Lol😂😂😂😂
That line killed me!!! XD XD XD
Ets genial . Des de Catalonia
"You guys should try and solve first"
Me at 3 am: No, no I dont think I will
Lol, it’s exactly what I did, just watch the thumbnail, was like « Ok (1;1) is a solution when x=y » and then... « If you multiply the second equation by 5 we get a polynomial expression that we can factorise by (x-y) and something of degree 2...damn I’ve got to try it » and so I did lol, it works fine.
8:46 golden ratio!!!!!!
6:30 instead of doing polynomial division, if you add y - y and split the -2y^2 into -y^2 and -y^2 you get (y^3 - y^2) - (y^2 - y) - (y + 1). You can now see that you can take out a (y-1) from each of those.
I got this, but admittedly I did do the polynomial division first then looked for a prettier solution once I knew the end result, but it’s still cool imo: A reminder that factoring non-obvious polynomials like this is in some ways about “creating or finding the symmetry” which I think is a helpful way to look at it for other factoring problems too.
3blue1brown gang!!!
Yea lol!!!
Plz do a collab on a topic of math plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
And Eddie Woo!
3brown1blue
The seamless switching of the pens was just as impressive as the math.
about the division made in 8:00, I can affirm that in Portugal, my country, we learn it in high school (10th grade), but it is called as Ruffini's Rule
6:32, I called this "Ruffini's rule", a particular version of "Horner's method" ( usually called synthetic division). I remembered I learned this in highschool, and I used to do it without thinking. Then I understood, how division of polynomials work, and never used it again until the point I completely forgot about the algorithm.
Yes it's a good way to crunch the numbers quickly in an exam but it's important to understand what's really going on behind the scenes
That's a golden answer for y.
Yeah thats omega golden ratio
.
that's phi not Omega@@justaconcrete4789
Shout out! "DO NOT Automatically substitute!" Haha especially differential equation. I like your pi. Would like to buy that one
The pi plushie is from 3blue1bown's store:
store.dftba.com/collections/3blue1brown?_=pf&pf_pt_product_type=Plushies#bc-sf-filter-products
@@scoutskylar but how did he make it into a microphone lol
@@mathadventuress Probably just a lapel mic clipped onto the back
This mic looks like a logo of one of the math youtube channel, but i don’t remember which exactly
@@equilibrium7756 3blue1Brown
This is my favourite content. These videos are keeping my math fresh while in quarantine. Thank you for making these!
You’re welcome! I am happy to hear : )
just like almost every single math problem u start losing urself throughout the process and question if 1 + (-1) is 0 :))
imagine you saw 9 as "a" and tried to solve problem like 1 hours ...
Wait it wasn't an A?!!!
@@Azmidium You can't imagine how i tried to solve this problem i was about to cry lmao xD
thanks for telling me... wasted 20 mins😔😭@@thesixteenthstudent2497
On 7:59, Bprp, you said "I'm not sure whether this is a popular method".
Yes, in Indonesia, the method (If I am not mistaken: Horner Method) was popular just for a while (maybe until 2013). But since Indonesian math curriculum did not add this in the newest curriculum (which is from 2014), not many know this method.
This video is very mathematically satisfying
Grant would be proud of your new microphone :)
First try: check if x=1, y=1
Result: it works 😳
Conclusion: I am a god.
Greetings from Poland 🇵🇱 (now, it's 3.57 AM here, so I should go sleep 😴).
siema byku
I also solved it that way
Wait I actually solved it using that
Same way I did
Guess and check ftw
What I did is that I first took x^3 common from the first equation and y^3 from the 2nd.Now I took (y/x) as some value 'a' and divided the two equations.When we simplify this we get a cubic equation having only the value 'a'. If u solve that equation, we get many more terms like the solutions of x^3 = 1 and y^3 = 1 and some more, which also includes the answers u found.But yours was an elegant method and much more easier. Great videos btw
Man I love the solutions you bring, they are always so fun to see!
Golden ratio sure appears everywhere.... including in olympiad problems
"Hopefully we can get points for this question" 😅👍 yes you do Steve. And I loveeeeeeee your little pi figure from 3blue1brown? And excellent video of course.
Yes. Thank you Sergio!
You can also divide equation 1 by equation 2 to reduce it to a cubic in terms of t=x/y however substituting back in to solve for x and y is much messier
Are you sure you performed the division correctly. How did you manage to get a cubic of x/y?
(1,1) is really easy to find just by inspection, especially if you factor out an x^2 and y^2 from each equation. Then you can skip straight to finding the other two solutions.
Idk about you guys but I just instantly saw 1 as a solution….
You can also put y=m*x in the 2nd eq and then divide both eqs
Glad to see 3b1b spreading around just as well as *the thing*
The system is symmetric with respect to the exchange of x and y. Then replacing y by x in either of the equations gives us x³=1. Then it is straightforward to find the three roots of x. The real solution is x=1, and the two complex roots are 120° symmetric with respect to the real root, on the complex plane.
0:25
Man talking to his dolls during isolation, 2020, colorized
BPRP’s videos are so good that it has 1 view and 7 likes
UA-cam algorithms!
Transdimensional maths.
Living in 4D
@@southernkatrina8161 Transdimensional Maths Yeah!
By inspection I noted that x = 1 and y = 1 are solutions.
I then factored out the solution for x or y as (x-1) or (y-1) and then solved the system for the other solutions.
Ahhh that's very clever! Finding (1,1) by inspection was not very difficult, but I forgot about factoring out those solutions to reduce each equation to a problem in quadratics. Excellent!
How do you factor these out?
@@Skandalos Using the division method probably. There are many methods but this is the method they teach in most schools.
Man, you use colours so well! A very efficient lecturer, too...
Salute to his highly developed brain and his courage to become so efficient in mathematics at such a young age
Everytime I see his vids along with the solutions I wonder why such solutions don't come to my mind?😂
"highly developed brain and his courage
to become so efficient in mathematics at such a young age" bro what?
bet nobody was expecting the 1,1 until they saw it and substituted it into the equation
Your approach is praiseworthy .
Great video. My expansions concept got cleared! Thanks!
I finally got one of these questions right!!! You guys don’t know how happy I am!!!
ua-cam.com/video/kow8ijXyVMQ/v-deo.html
Press f for those who tried substitution before watching the video 😂
F
F(1)
I tried 3 different subtractions of these equations, only to find that the solution was a sum... (facepalm).
By the way, what does this "press f" cliche mean ?
Press f to pay repesct
what a nice video!, youtube algorithm making my quarentine night better
Hahaha thank you.
Nice! The solution presented in this video is super elegant!
Here's a nice trick :
Multiply the bottom equation by 5 and subtract it from the first equation. The resulting equation is equal to zero.
Now notice x = y is a root of the equation since the sum of coefficients is zero.
Divide out the factor (x-y) using long division. Now solve for the other factors using the quadratic formula!
Edit : Don't forget to substitute into either of the first two equations to find the 'initial condition' of the system
Lol I just thought of same thing
Could you elaborate on this please?
(x-y)(x^2 + 10xy + 5y^2)=0. Is that what you meant by dividing out the factor
(x-y) from the equation? If so I don't see how I can apply the quadratic formula or how it helps me, I would be very grateful for your assistance.
On further reflection using your method I get y= cuberoot(1/(-2 +- sqrt(5)) which gives the right answer but not in the same form.
@Memes shorts Thank you.
@Memes shorts Well UA-cam informed me of your reply, so I wanted to thank you for taking the trouble. You're right it is amusing though 😂.
Here's an interesting problem:
Find all integers "x" that satisfy x/a = x-a for a given integer "a"
Rearranging gives us a^2-ax+x=0
We will search for integer values of x where the solution a is an integer too. a=(x+-sqrt(x^2-4x))/2
For a to be integer we should have no square roots, thus x^2-4x must be a perfect square. x^2-4x can be written as (x-2)^2-4, which is some perfect square minus 4. Now this number is only a perfect square when it is equal to zero (can someone help me prove this?)
x^2-4x=0 gives x=0 and x=4
Quickly checking gives us for x=0 a=0 and for x=4 a=2
I dont think I missed any solutions, I just need to prove that (x-2)^2-4 is only a perfect square when equal to zero, can someone tell me how to do it?
Edit - nvm I figured it out, here you go:
The question is: When is x(x-4)=n^2 for some integers x and n
(n larger than zero)
we quickly notice that for this to be the case, our 2 factors must be related to n,one of them must be divisible by n and the other must be a divisor of n
, the larger should be divisible by n, so x is divisible by n and x-4 is a divisor of n by the same coefficient:
meaning that x=bn for some integer b and n=b(x-4) for the same value of b. substituting n in the first equation:
x=b[b(x-4)]=b^2(x-4)
x=(b^2)x-4b^2 thus x(b^2-1)=4b^2
x=(4b^2)/(b^2-1)
we can factor b^2-1 into (b-1)(b+1)
and 4b^2 is (2b)^2
since x=((2b)^2)/(b-1)(b+1)
and since x is an integer
then (2b)^2 is divisible by both b-1 and b+1
using the logic we used earlier b+1=c(2b)
and b-1=(2b)/c
b+1=c^2(b-1)
b(c^2-1)=c^2+1
b=(c^2+1)/(c^2-1)
b=((c^2-1)+2)/(c^2-1)
b=1+(2/(c^2-1))
since b is an integer and 1 is an integer, 2/(c^2-1) should also be an integer. c^2-1 has a minimum value of -1, which gives b=-1, and b=-1 gives no x solutions. For c=1 b is undefined (1+2/0)
, for c greater than or equal to 2, c^2-1 is greater than 2, denominator greater than numerator aka 2/(c^2-1) wouldn't be an integer. Thus b can't be an integer and by contradiction there are no solutions (other than the case where n=0, which gives x=0 or x=4)
One more thing, x=0 and a=0 is not a valid solution, since 0/0 is undefined. Thus the only solution we have is x=4 a=2
Edit 2 - I just noticed that c isn't necessarily an integer, but c^2 is. Solution isn't affected much though, no new x values appear.
Edit 3 - Apparently b^2 and b do not have to be integers, but c^2 has to, so we can just substitute integer values of c^2 and check if we get a valid x solution. If anyone has a more efficient way of doing this please tell me.
Nevermind I found a much easier solution. Since both x and a are integers, x-a is an integer, thus x must be divisible by a. Let x=an for some integer n, we have:
an/a=an-a
n=an-a
n=a(n-1)
a=n/n-1
Now we need a to be an integer, so we can try some n values.
For n=0 a=0 (rejected)
For n=1 a=1/0 (rejected)
For n=2 a=2, which gives x=4 (accepted solution)
For n above 2, n-1 becomes greater than 1, n and n-1 would be relatively prime, and thus we wouldn't be able to divide n by n-1 and get an integer.
For negative values of n,
n=-1 a=1/2
n=-2 a=2/3
Etc
Thus the only solution is the pair x=4 and a=2
You can intuitively tell it's 4 and 2 just looking at it. What other pair of integers has the same difference and quotient
If there is a collab betwem bprp and 3b1b, little rabbit and pieeeeee will be very happy 😍
Loving the new microphone!
Well your equation has a golden solution
At y^3-2y^2+1=0, first thing I notice is that the coefficients add up to 0, if they add to 0 then (y-1) is a factor.
This night I weirdly dreamt about watching a long bprp video and forty minutes in we start hearing violin and he's kinda mad since it's his children and they are not practicing the piece they're supposed to. Given that Steve seemed the same as he is now, his children were likely very young while playing at professional level. Weird dream, but fun dream.
Don’t forget the complex solutions by multiplying x and y by a third root of unity
noahtaul He said specifically that they are only looking for real numbers.
About that dividing polynomials method you used in 8:00 what i can confirm is in Poland where i live about 10 years ago while i was in high school we had it but it was class with advanced math profile, classes with basic math didn't have it and last time i checked like 2-3 years ago advanced classes didn't have that method anymore
Are x and y real numbers?(I couldn't find the conditions). If complex numbers are also possible, x=y=(-1+sqrt(3)*i) / 2 and x=y=(-1-sqrt(3)*i) / 2 can be solutions as well.
Красиво. Особенно разложение на множители. Разделил оба на xy^2, замена x/y=u, доумножил на -5, сложил, получил кубическое уравнение u^3+4u-5=0. Дальше, только разглядев u=1, смог разложить на множители, без подсказки видео не додумался. Здоровское решение, спасибо.
since the given equations are homogenous in nature put y=kx, later eliminate x^3 find values of k, we will get one solution as k=1 and another quadratic equation of k, let the roots be k1,k2,k3 we will get y1,y2,y3, and substitute in the given equations, we will get x1,x2,x3 now again put in y1=k1*x1, similarly y2=k2*x2, and y3 also
What you got are full real solutions of the problem,
but if we consider the complex number field, I think that there can be three cases in 4:21
As you know in complex field, (x+3y)^3 = 4^3 has three cases.
x+3y = 4 or x+3y = 4e^{2pi i/3} or x+3y = 4e^{4pi i / 3}
I don't know full statement of that problem in original olympiad contest.
But if we want to find full 'complex' solutions, it would be troublesome!
I solved it within a 10 sec x=1 and y= 1
Nicely done sir! Really like your videos!!!
Love your show. Now you need a bigger board so you won't have to erase the first half to present the second.
x=1 and y=1 just by observation
x³ + 9x²y = 10
y³ + xy² = 2
First blush of my thought process:
Hmmm..the LHS of the 1st equation looks like the start of the expansion of (x + 3y)³ - what's the rest?
Let's see, x³ + 3·3x²y + 3·9xy² + 1·27y³ = x³ + 9x²y + 27xy² + 27y³ = x³ + 9x²y + 27(xy² + y³)
And son-of-a-gun! There's the LHS of the 2nd equation!
So
(x + 3y)³ = 10 + 27·2 = 64
x + 3y = 4 [or 4 times either of the two complex roots of unity]
Hey! How about x = y = 1? YES!! That works! That could have been guessed without going through any of that rigamarole above.
What about other possible solutions, even without using the complex roots we set aside earlier?
y = ⅓(4-x)
10 = x³ + 3x²(4-x) = -2x³ + 12x²
x³ - 6x² + 5 = 0
ffggddss how can you do exponents and fractions in this comment box
2x³+15x²y=25
5y³+6xy²=1
Then x=? y=? Challenge for you
@@midorimashintaro2613 Some are available on my Mac keyboard using Option and Option-Shift (in Windows, "alt" & "alt-Shift");
some I've collected from other posts on Yahoo Answers & YT, by Copy-Paste into a text file, then from there to YT comments.
In Windows, I understand there are ways to get these Unicode characters using the keyboard...
Fred
It looks give more questions than answers.
How to calculate the number of roots/solutions for this equation (it looks like having total 5 roots in WFA)?
Why can you divide the equations (1) and (2) with (x-1)*(y-1) and then solve rest of the roots?
Since both equations are homogenous. Let y = mx for some real number m and substitute that in both equations. This will eliminate the y variable. Now factor x^3 from both equations and divide the equations . Now youll get the cubic polynomial in m . Solve for m and back substitute to get values of x & y for corresponding values of m.
Is that the 3b1b pi? I gotta get one of those
"..... why do I want to do that? No. I am blackpenredpen...." loved it and the new mic !
1:52 LITTERAL CHILLS
Blackpenredpen did u can solve the question in by like 2 seconds by using cross method multiplication. (10*1-9*2)/1*1-9*1= - 8/-8 = 1 thus the first row =1 therefore x and y equal 1(the row only works in this case because of x and y having no coefficient usually it is only x and u sub in to get y)
好特別的題目，謝謝老師
: )
謝謝收看！
blackpenredpen 一定的！
these are homogeneous equation .can be solved by taking commom x^3 and y^3v respectively from both equations and then divide.
y3 - 2y2 + 1 =0
y3 - y2 -y2 + 1 =0
(y3 - y2) - (y2 - 1) = 0
y2(y - 1) - (y-1)(y+1)=0
(y-1)(y2 - (y+1)) =0
(y-1)(y2 - y -1) =0
Yes, y=1 is a solution. The other two solutions are irrational.
Do i get any points for just guessing the x=1m y=1 solution at the start?
5:01 I thought you would rewrite it as 3y = 4 - x and substitute it into the first equation. lol
Thanks teacher from Brazil!
Funny thing is, I "substituted right away" x = 1, y = 1. Another attack is to divide the first equation by the second to get (t^3 + 9t^2)/(1+t) = 5 where t = x/y. So t^3 + 9t^2 - t - 5 = 0 where we know t-1 divides the LHS since the solution x=1, y=1 gives t = 1 as a root. The other two solutions (after synthetic division, quadratic formula) are t = -5 + 2\sqrt{5}, t = -5 - 2\sqrt{5}. It's not too hard from there. Also one shouldn't worry about real versus complex: by homogeneity of the left-hand sides, if (x, y) is a solution, then so is (cx, cy) where c is a cube root of unity.
1:10 bprp "blue pen red pen yey!"
Solved it immediately by guessing a solution.
You make maths fun.
Epic video, thank you!
8:00 "I'm not sure this is the popular method"
Popular? In middle & high school every student hated Ruffini's rule ahah.
But yes, it's a good quick method I think
Glad to see a 3Blue1Brown Pi mic
Solving the second equation for x, and substituting into the first equation is not very difficult. You get a cubic in y^3, and y^3=1 is one of the solutions. Divide it out, and you get a quadratic in y^3. If I didn't make a mistake, y^3 = 2 +- sqrt(5) is the other pair of solutions. x = (2 - y^3)/y^2 -- Now getting all pairs (x,y) might be a bit more grungy.
Very easy for me
Thank you
Finally set of equations with y = Φ.
There are 6 more non-real solutions
Yeah, probably, since he only took the real third root on purpose
Just addicted to your channel
Man how many toys do you have ?
Nice 3b1b pi plushie!
Sir please make a series video on indefinite integration from basic to advance
search "ultimate integral starter"
aye!
3b1b and bprp Collab
nice
That's a very cool way to solve this system. I multiplied the bottom one by 5 and then subtracted it from the top one to get a homogeneous equation, however it gets kinda messy after solving for x/y. It's not that hard, but that's a definitely more arithmetic involved approach. It would be nice to see more geometry on your channel as well, btw
I did the same. Once i saw how he did it i felt like an idiot lol
What is connection with homogene equation? There is no derivates
(Sorry for bad english)
@@rahimeozsoy4244 well, a function is called homogeneous if it has this property: f(ax)=(a^n)f(x), and n is called a degree of homogeneity. It's true for homogeneous differential equations as well but that's not the only case. For any homogeneous equation of two variables it is true that dividing equation by y^n (or x^n, it's symmetric) you will get a polynomial of degree n with respect to x/y (or y/x). Any monomial is homogeneous (it's rather obvious) and every polinomial that consists of monomials of same degree is homogeneous too (for that it should not include any nonzero constants). So in a sense, calling something homogeneous means that this function/equation is symmetrical (loosely speaking) upon change of x and y and proportional too their multiples. I like to think about it as it means that some sort of equivalency in geometry (homothety). It's equal to proportion. So having everything in terms of x/y (y/x) makes sense, you are concerned only with ratios, not actual values, and after figuring out ratios you can easily (in a mathematical sense) figure out values. That's a brief overview, but I recommend to read a little bit about homogeneous functions on your own, it's not a particularly difficult topic.
Oh, and there's no need to apologise for any mistakes, English isn't my native language as well, as suggested by my name :), and even if it was it's not a big deal.
@@leonhardeuler6811
Looks like your skills got rusty after all these years, Leonard :)
@@vladanikin696 Where I search homogene equation I always find out differantial homogene equation. I thought homogene is connected with derivates.
Adding Eq1 + 3 x Eq2 gives (x + 3y)^3 = 64 = 4^3. So, x + 3y = 4 x (1, w, w^2) [ w = cube-root of unity ]
In 1st equation just add & subtract X^2Y & XY take common (X-1)
X-1=0
X=1 put in equation Y= 1
Cool solution! Alas... I would have never figured it out. So congratulations are in order ... 👍
isn’t it just (1,1)?
@5:08 I managed to deduce in my head that (1,1) was a solutionl
Thx for sharing
I would rather multiply the second equation with 5 and then eliminate the constants by subtracting equation. Then divide the formula by y³ and you get a cubic equation in q=x/y. Then you solve for q, and plugin x=qy in any of the given equation and find x and y. I think this is better because if you get uglier constants it will work
There are 9 solutions
x'=cos(2π/3), y'=isin(2π/3) and
x"=cos(4π/3), y"=isin(4π/3) are olso other solutions for (1,1).
for every solutions (including imaginary numbers) just rotate each solution by 2π/3 and 4π/3 {it means multiply e^(2πi/3) and e^(4πi/3)}
I was trying to solve this for 3 months with no success....and it was so simply after I've watched you...
Nice!
3b1b reference mic.
we want more national Olympiad Problems!