4
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does $a^2-51b^2=\mp 6$ have a solution for integers?

I have tried for many modulos, but could not get much out of them.

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  • $\begingroup$ Are you familiar with quadratic residues? $\endgroup$ – Daniel Fischer Dec 24 '13 at 21:00
  • $\begingroup$ yes. would mod 5 work? $\endgroup$ – 104078 Dec 24 '13 at 21:02
  • $\begingroup$ The key, incidentally, is to pick a modulus that simplifies the problem; since $51=3\cdot17$, those are the places to start. $\endgroup$ – Steven Stadnicki Dec 25 '13 at 1:36
  • $\begingroup$ Related : math.stackexchange.com/questions/603104/… $\endgroup$ – lab bhattacharjee Dec 25 '13 at 3:27
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A solution to $a^2 - 51 b^2 = \pm 6$ would in particular be a solution to the congruence

$$a^2 \equiv \pm 6 \pmod{17}.$$

But neither $6$ nor $11$ is a quadratic residue modulo $17$ - the quadratic residues are $1,2,4,8,9,13,15,16$, so there is no solution.

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  • $\begingroup$ beat me to it! :) $\endgroup$ – frogeyedpeas Dec 24 '13 at 21:04
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$\rm mod\ 17\!:\ a^2\!\equiv \pm 6\Rightarrow a^4\!\equiv 6^2\!\equiv 2\Rightarrow a^8\!\equiv 4\Rightarrow a^{16}\!\equiv -1\,$ contra little Fermat. $\ $ QED

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