I deduced that $8z^2+1=(8x-1)(8y-1)$, but then I don't know what to do.
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$\begingroup$ Wolfram Alpha suggests there are no solutions where $x,y,z$ are all positive. $\endgroup$ – Toby Mak Jul 31 at 2:48
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$\begingroup$ math.stackexchange.com/questions/3214034/… $\endgroup$ – individ Jul 31 at 4:23
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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
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$\begingroup$ 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? $\endgroup$ – JavaMan Jul 31 at 2:54
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1$\begingroup$ @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 $\endgroup$ – Will Jagy Jul 31 at 2:59
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$\begingroup$ Ah. I missed that the primes $q \equiv 5, 7 \pmod{8}$ had to be positive of course. $\endgroup$ – JavaMan Jul 31 at 3:00
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$\begingroup$ @JavaMan I ran a short script and added some (negativexy)solutions to comment above $\endgroup$ – Will Jagy Jul 31 at 3:01
