Integral solutions to $A^2=(u^2+uv+2)/2$, $B^2=(v^2+uv+2)/2$, $C^2=1+uv$

Question

The system of equations $A^2=(u^2+uv+2)/2$, $B^2=(v^2+uv+2)/2$, $C^2=1+uv$ has $A,B,C,u,v$ all integral. Numerical evidence found by brute force computation implies that $uv(u^2-14uv+v^2-16)=0$. Certainly solving this equation gives an infinite number of solutions so it is sufficient, however I am unable to prove it is necessary. (Solving involves using either $X^2-2Y^2=1$ or $X^2-3Y^2=1$ and standard Pell techniques.) So can anyone help show that ALL solutions are given by $uv(u^2-14uv+v^2-16)=0$

Edit

This question is a formulation of Problem 10622 in American Mathematical Monthly 1997 p870. The problem only asks for an infinite number of solutions, I want to find all of them.

My approach is first to note that $0\le C^2=1+uv$ implies $uv\ge-1$.

$uv=0$ gives if $u=v=0$, $A^2=B^2=C^2=1$ as solutions. $u=0,v\ne0$ gives $A^2=C^2=1$ and$$2B^2-v^2=2$$noting that this implies $2\mid v$ then gives$$B^2-2(\frac{v}{2})^2=1$$and standard Pell techniques gives all solutions. Obviously $u\ne0,v=0$ is similar. This provides all solutions to the $uv=0$ part of $uv(u^2-14uv+v^2-16=0$.

$uv=-1$ gives $C=0$ and $u=\pm1,v=\mp1$ hence $A^2=B^2=1$ as solutions. In this case $u^2-14uv+v^2=16$.

Now noting that $u$ and $v$ cannot have opposite signs (except $uv=-1$ above) and the symmetry of the system of equations means we can limit to $u,v>0$ and change signs to get a complete set. The symmetry also means that we can assume $0<u\le v$ wlog.

$u=v$ implies $C^2=1+u^2$ which gives $u=v=0$ which has already been covered. Hence we can now just look at $0<u<v$ and show that necessarily $u^2-14uv+v^2=16$.

1. For all $u,v$ we have $u^2-14uv+v^2\equiv0\pmod{16}$. This can be shown, if need be, by looking at all 256 possibilities.

2. For all $u,v$,$$A^2=(\frac{v-3u}{4})^2-(\frac{u^2-14uv+v^2-16}{16})$$$$B^2=(\frac{3v-u}{4})^2-(\frac{u^2-14uv+v^2-16}{16})$$$$C^2=(\frac{u+v}{4})^2-(\frac{u^2-14uv+v^2-16}{16}),$$which show that $u^2-14uv+v^2=16$ is sufficient.

I have found no other way forward that doesn't simply provide tautologies or dead ends. Hopefully this will all help someone else find some inspiration!

Answer 1

I thought to find out why $x^2 - 14 xy + y^2$ is involved, perhaps this is the source of the problem: if $$ x^2 - 14 xy+ y^2 = 16 - \delta $$ we find $$ 16(1 + xy) = \delta + (x+y)^2 $$ $$ 8(x^2 + xy + 2) = \delta + (3x-y)^2 $$ $$ 8(y^2 + xy + 2) = \delta + (-x+3y)^2 $$

Which leads us to ask: what can we say when $ \; r^2 + \delta, \; s^2 + \delta, \; t^2 + \delta \; , \; \;$ are all squares? Here $\delta$ may be positive or negative. Maybe just inequalities to continue

next thought. $x+y$ is the average of $3x-y$ and $-x+3y$

Well, not that restrictive, $13^2 + 120 $ and $7^2 + 120$ and $1^2 + 120$ are all squares. Then
$27^2 -504 $ and $25^2 -504$ and $23^2 -504$ are all squares.