Find all positive integers $a,b,c,d,e,f$ such that :
$de^2=ab^2+1$ and $df^2=ac^2+1$.
I tried subtracting them, it factors quite nicely. But after that, haven't a clue. I'm not sure if it's even possible to prove in general. I have managed to transform it into various Pell's equations by e.g. letting one of the terms be $1$ and so on, but these miss the big picture. Does a general characterization exists that finds all solutions?
Thanks very much for your help!
If the quadratic form $d x^2 - a y^2$ integrally represents $1,$ then it does so infinitely many times, all of which can be found by constructing the full automorphism group. The first step is to find the fundamental (first nontrivial) solution to $$ u^2 - 4ad v^2 = 4; $$ note that $u$ will be even, so we actually have $$ w^2 - a d v^2 = 1. $$ Then the matrix that (with its inverse) generates the group is $$ \left( \begin{array}{rr} w & a v \\ d v & w \end{array} \right) $$ See how the determinant is positive $1.$
If you have a solution $(x,y),$ you get a new solution with $$ (wx + avy, \; dv x + wy). $$ There is always the reflected solution $(x,-y)$ as well.
Meanwhile, many such diagonal indefinite forms simply do not represent $1,$ despite what appear to be favorable circumstances $\mod a$ and $\bmod d:$
Here is a good one, $2 x^2 - 41 y^2,$ does not represent $1$ or $-1.$
For discriminant $328,$ the principal genus has two forms, $$ \langle 1, 0, -82 \rangle, \; \; \langle 2, 0, -41 \rangle, $$ while the other genus has two opposite forms, $$ \langle 3, 2, -27 \rangle, \; \; \langle 3, -2, -27 \rangle. $$