For prime numbers $p_n$ it holds $m\equiv p_n^2\pmod {p_{n+1}}\implies$ $m$ is a square

Question

Conjecture about primes:

For all $n\notin\{4,9,30\}$ it holds that if $m\equiv p_n^2\pmod {p_{n+1}}$ where $0 \leq m < p_{n+1}$, then $m$ is a square.

Tested for $n<1,000,000$.

The only exceptions seems to be $p_4=7$, $p_9=23$ and $p_{30}=113$.

Would like help to prove, explain or contradict this.

Answer 1

This is true if either $p_n^2<p_{n+1}$ (obviously) or if $(p_{n+1}-p_n)^2<p_{n+1}$ (because these two squares are congruent mod $p_{n+1}$, since $(-1)^2\equiv 1$). Since a number only has two square roots (or none) modulo a prime, these are the only possibilities.

So the statement is equivalent to the gaps between consecutive primes being less than the square root of the higher prime. This being true in general is basically Oppermann's conjecture (what we need is very slightly stronger, see comments), so it's a deep unsolved problem.