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

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closed as off-topic by user21820, callculus, vonbrand, GNUSupporter 8964民主女神 地下教會, The Count Aug 9 at 0:46

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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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    $\begingroup$ @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
  • $\begingroup$ 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
  • $\begingroup$ @JavaMan I ran a short script and added some (negativexy)solutions to comment above $\endgroup$ – Will Jagy Jul 31 at 3:01

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