find all p, for which the determinant is divisible by $p^{3}$

Question

So, we have to find all prime numbers $p$ such that

Answer 1

Hint: One approach for getting an expression for the determinant is to use the matrix-determinant lemma with $$ A = \pmatrix{2^2 - 1\\ & 3^2-1 \\ && \ddots \\&&& (p+7)^2 - 1}, \quad u = v = (1,1,\dots,1)^T. $$ Doing so yields the expression $$ \det(M) = \left(1 + \sum_{i=2}^{p+7} \frac{1}{i^2 - 1} \right)\cdot \prod_{i=2}^{p+7} (i^2 - 1)\\ = \left(1 + \frac 12 \sum_{i=2}^{p+7} \left[\frac{1}{i-1} - \frac{1}{i+1}\right] \right)\cdot \prod_{i=2}^{p+7} (i^2 - 1). $$ Note that the sum is telescoping.