# All even squares are congruent with a prime square $p_n^2\pmod {p_{n+1}}$

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 $m\leq 152$.

• 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. – Mr Pie Feb 10 '18 at 8:13
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 $p_n > m$.