Does the equation $$z(y-x)(x+y)=x^3$$have finitely many solutions in the set of positive integers? Proposed by Nikola Velov
Problem
Source: 4th Memorial Mathematical Competition "Aleksandar Blazhevski - Cane"- Junior D2 P4
Tags: number theory, Junior, construction
hi_how_are_u
21.12.2022 04:11
pick $y = k*x+x$
$z*(kx)(kx+2x)=x^3$
$z*k*(k+2)x^2=x^3$
$x=z*k(k+2)$
$y=x(k+1) = (z)(k)(k+2)(k+1)$
for any k and z thus the equation has infinitely many solution since (k,z) could be any natural
to simplify more the solution we can also pick
$z=k$
$x=k^2(k+2)$
$y = k^2(k+2)(k+1)$
which is work for any $k \in \mathbb{N^{+}}$
physicskiddo
21.12.2022 04:59
hi_how_are_u wrote:
pick $y = k*x+x$
$z*(kx)(kx+2x)=x^3$
$z*k*(k+2)x^2=x^3$
$x=z*k(k+2)$
$y=x(k+1) = (z)(k)(k+2)(k+1)$
for any k and z thus the equation has infinitely many solution since (k,z) could be any natural
to simplify more the solution we can also pick
$z=k$
$x=k^2(k+2)$
$y = k^2(k+2)(k+1)$
which is work for any $k \in \mathbb{N^{+}}$
Nice, brilliant sol
nsd08
21.12.2022 05:34
Wow. How do you think to do something like this?
nsd08
21.12.2022 05:39
i am too tired to do math
physicskiddo
21.12.2022 05:45
nsd08 wrote:
set $y=0$
$-zx^2=x^3$
$-z=x$
so there are an infinite number of solutions. this obviously isn't as rigorous as the above sol
Unfortunately I don't think this works because $0$ isn't a positive integer
Jt.-.
21.12.2022 05:46
$z(y^2-x^2)=x^3$
$zy^2=x^2(x+z)$
$y^2=\frac{x^2(x+z)}{z}$
We want $\frac{x+z}{z}$ to be a square, so just make it $4$!
Then we get $x=3z, y=6z, z=z$ which works, so we are done.
physicskiddo
21.12.2022 05:46
nsd08 wrote: Wow. How do you think to do something like this? I am also curious about this. I think the first idea someone might have is eliminating some form of nasty term... having $y=f(k) + x$ and from there trying things might work. But it still seems like some sort of superpower to me.
IAmTheHazard
21.12.2022 06:21
Let $z=1$. The equation then becomes $y^2=x^3+x^2=x^2(x+1)$. Then just take $x$ such that $x+1$ is a square. The other dumb thing to do is to just note that if $(x,y,z)$ is a solution then $(xk,yk,zk)$ is as well, so find an explicit solution (not hard) and then use that to generate infinitely many...