Solve in positive integers $8xy-x-y=z^2$ [closed]

I deduced that $$8z^2+1=(8x-1)(8y-1)$$, but then I don't know what to do.

the problem is the positivity restriction. When $$x,y$$ are positive, both $$8x-1$$ and $$8y-1$$ are $$7 \pmod 8.$$ As a result, each fails to be a product of (positive) primes $$1,3 \pmod 8.$$ Each is divisible by at least one (positive) prime $$q \equiv 5,7 \pmod 8.$$ However, any $$1 + 2 u^2,$$ including your $$1 + 8 z^2,$$ can only be divisible by primes $$p \equiv 1,3 \pmod 8.$$

I guess i should specify the quadratic form lemma, people don't seem to know this: Lemma. If $$v^2 + 2 u^2$$ is divisible by some (positive) prime $$q \equiv 5,7 \pmod 8,$$ then both $$u,v$$ are divisible by $$q.$$ In particular, $$v \neq 1.$$

here are some solutions with negative $$x,y$$

 x:  -1 y:  0 z:  1
x:  -5 y:  -4 z:  13
x:  -7 y:  -1 z:  8
x:  -7 y:  -2 z:  11
x:  -11 y:  -1 z:  10
x:  -17 y:  -16 z:  47
x:  -19 y:  -5 z:  28
x:  -25 y:  -19 z:  62
x:  -29 y:  -4 z:  31
x:  -31 y:  -2 z:  23
x:  -32 y:  -1 z:  17
x:  -37 y:  -4 z:  35
x:  -37 y:  -7 z:  46
x:  -40 y:  -1 z:  19
x:  -45 y:  -31 z:  106
x:  -49 y:  -39 z:  124
x:  -54 y:  -19 z:  91

• Why does the positivity requirement cause a problem here? It seems if $x, y$ are any integers, then $8x - 1$ and $8y - 1$ are $7 \pmod{8}$, so what changes when $x$ and $y$ are negative? – JavaMan Jul 31 at 2:54
• @JavaMan then the absolute values of the two factors are $1 \pmod 8.$ There are many, many such solutions. Same effect as changing the problem to $(8s+1)(8t+1) = 1 + 8 r^2$ x: 0 y: 0 z: 0 x: -1 y: 0 z: 1 x: -2 y: -2 z: 6 x: -4 y: 0 z: 2 x: -5 y: -4 z: 13 x: -7 y: -1 z: 8 x: -7 y: -2 z: 11 x: -9 y: 0 z: 3 – Will Jagy Jul 31 at 2:59
• Ah. I missed that the primes $q \equiv 5, 7 \pmod{8}$ had to be positive of course. – JavaMan Jul 31 at 3:00
• @JavaMan I ran a short script and added some (negativexy)solutions to comment above – Will Jagy Jul 31 at 3:01