There was one task on the competition http://kskedlaya.org/putnam-archive/
I'm not much will change. Is it possible to solve such a system of Diophantine equations?
$$x(x-a)+2yz=y(y-a)+2xz=z(z-a)+2xy$$
$a - $ The specified number for the problem. I think we should first find a parametrization of the solutions of this system of equations taking $a$ as unknown. And then to find out for which values of solutions are possible?
On
In General, the problem looks simple. And I don't understand why. This equation can be rewritten in such a system.
$$\left\{\begin{aligned}&x(x-a)+2yz=y(y-a)+2xz\\&z(z-a)+2xy=y(y-a)+2xz\end{aligned}\right.$$
This is equivalent.
$$\left\{\begin{aligned}&(x-y)(x+y-a)=2z(x-y)\\&(z-y)(z+y-a)=2x(z-y)\end{aligned}\right.$$
This is equivalent.
$$\left\{\begin{aligned}&x+y-a=2z\\&z+y-a=2x\end{aligned}\right.$$
This will mean that:
$$x=z=y-a$$
Or maybe somewhere made a mistake?
Since the terms involving $a$ are all homogenous of degree one, and the other terms are all homogenous of degree two, the parameter $a$ can be removed by a simple rescaling: set $x=ax'$, $y=ay'$, $z=az'$ and the equation becomes $ax'(ax'-a)+2ay'az'=$ etc, where we can factor out to get $a^2(x'(x'-1)+y'z')=$ etc. In other words, $a$ is a 'false parameter'; the equations are no more general for its introduction than they were without it (n.b.: the original problem from the Putnam exam is the case $a=1$), since solutions of one set map directly to solutions of the other.