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A conjecture about primes related to the question
For prime numbers $p_n$ it holds $m\equiv p_n^2\pmod {p_{n+1}}\implies$ $m$ is a square

For every even $m>0$ there is a $n\in \mathbb Z_+$ such that $m^2\equiv p_n^2\pmod {p_{n+1}}$, $0\leq m^2<p_{n+1}$.

Tested for $m\leq 152$.

Are there some heuristics about this? Or counterexamples?

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  • $\begingroup$ This is almost the same as each even number occurring as a prime gap. $\endgroup$ – Lord Shark the Unknown Feb 10 '18 at 8:00
  • $\begingroup$ @LordSharktheUnknown: How come? $\endgroup$ – Lehs Feb 10 '18 at 8:04
  • $\begingroup$ He said almost... $\endgroup$ – Mr Pie Feb 10 '18 at 8:08
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    $\begingroup$ If $$m^2\equiv p_n^2\pmod{p_{n+1}}$$ then $$p_{n+1}\mid (p_n + m)(p_n - m).$$ Since $m$ is even and $m > 0$, there is a chance that $p_n \pm m$ is a prime, particularly $$p_{n+1} = p_n\pm m.$$ Of course though $p_n < p_{n+1}$ so $p_{n+1} - p_n > 0$ and therefore we consider $$p_{n+1} = p_n + m$$ or $$p_{n+1} - p_n = m.$$ This is what @LordSharktheUnknown means by his comment, but perhaps $p_n + m$ is not prime, or not equal to $p_{n+1}$, so we add an almost to the statement. $\endgroup$ – Mr Pie Feb 10 '18 at 8:13
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    $\begingroup$ @user477343: It might happen, if my conjectures are random enough. ;) $\endgroup$ – Lehs Feb 10 '18 at 11:02
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If $$m^2\equiv p_n^2\pmod{p_{n+1}}$$ then $$p_{n+1}\mid (p_n + m)(p_n - m).$$ Since $m$ is even and $m > 0$, there is a chance that $p_n \pm m$ is a prime, particularly $$p_{n+1} = p_n\pm m.$$ Of course though $p_n < p_{n+1}$ so $p_{n+1} - p_n > 0$ and therefore we consider $$p_{n+1} = p_n + m$$ or $$p_{n+1} - p_n = m.$$ So if we can prove that every even number is the difference of two adjacent primes, we prove your conjecture. Also, $p_n - m > 0$ and therefore $p_n > m$.

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