parmenides51 wrote: Prove that there are infinite triples of positive integers $(x,y,z)$ such that $$x^2+y^2+z^2+xy+yz+zx=6xyz.$$ Vieta Jumping. Start from $(1,1,1)$, which works. Now, if $x \leqslant y,z$, do $x \to 6yz-y-z-x$, which is also root of $LHS-RHS$ as function of $x$. Note that $6yz-y-z-x>4yz-y-z-x \geqslant 2yz-x \geqslant x$, so it's works.

Similar to Bulgaria Junior winter 2020. Greedily set $z=1$, we want $x^2 + y^2 + 1 + x + y = 5xy$, i.e. $x^2 - x(5y-1) + y^2 + y + 1 = 0$. Discriminant is $D = (5y-1)^2 - 4y^2 - 4y - 4 = 21y^2 - 14y - 3$ which is odd so if it is $m^2$ for odd $m > 0$, then $x = \frac{5y-1+m}{2}$ is a positive integer. So it suffices to show that $21y^2 - 14y - 3 = m^2$ has infinitely many positive integer solutions. Transforming this info a Pell equation is enough.

Assassino9931 wrote: Similar to Bulgaria Junior winter 2020. Where can we find the statement of that problem?

Achilleas wrote: Assassino9931 wrote: Similar to Bulgaria Junior winter 2020. Where can we find the statement of that problem? It's not publicly available, so I'll just state it here - show that $2a^2 + 2b^2 + 2c^2 + 2d^2 = 3abcd$ has an integral solution with $a>2020$.

$x^{2}+y^{2}+z^{2}+xy+yz+xz=6xyz$ Multiply both sides by 2 and after grouping and completing squares we get, $(x+y)^{2}+(z+y)^{2}+(x+z)^{2}=12xyz$ Now multiplying both sides by $k^{2}$ where k is any positive integer $(kx+ky)^{2}+(kz+ky)^{2}+(kx+kz)^{2}=12xyzk^{2}$ For every value of k this equation will be satisfied and since k has infintie possible values this equtaion will have infinite solutions. $\boxed{Hence proved}$

I will just write the simple path of the solution: $(1,1,1)$ follows the equation. if we put y=1 and z=1, x=3 comes up. we do this all over again then, we end up with an infinite number of $(x,y,z)$s.