I've heard that if a number $$$n$$$ has quadratic surplus under a odd prime $$$p$$$ so it has around 50% accuracy to be a perfect square number.

Is it correct? How to prove?

If so,perhaps there exists a method to judge whether a big number $$$n$$$ is a perfect square number or not is that random amount of odd primes which are not divisors of $$$n$$$ and find Quadratic surplus of $$$n$$$ mod $$$p$$$. Failure appears means $$$n$$$ isn't a perfect square number.

I met a problem recently which need to prove the below conclusion.

There is a example that $$$n=13,p=5,2p+1=11$$$ meet this constraint

I wrote something and check it's correct when $$$40\le n\le 10^5$$$ but is there any complete proof for such B exists?

I've learnt Bertrand-Chebyshev Theorem that there exists a prime $$$p\in[n,2n]$$$ and it may help.

Thank you.