Distinct Sylow P subgroups

The question is: If the number of Sylow $p$ groups is not congruent to $1 \pmod{p^{2}}$, then there exists two distinct Sylow $p$ subgroups $P,Q$ such that $[P : P\cap Q] = [Q : P\cap Q] = p$. I know that whenever $PQ$ is a group, $|PQ| = \frac{|P||Q|}{|P \cap Q| }$ and that there are $1+kp : 1 \leq k \leq p-1$ possible Sylow $P$ subgroups, but I do not know how to proceed.

This is a special case of a stronger version of the usual Sylow Theorem; it appears as Theorem 1.16 in Martin Isaacs Finite Group Theory (AMS, Graduate Studies in Mathematics, Volume 92).

Given a finite group $G$ and a prime $p$, let $n_p(G)$ be the number of Sylow $p$-subgroups of $G$. The usual Sylow Theorem shows that $n_p(G)$ divides $|G|$, and $n_p(G)\equiv 1\pmod{p}$. But in fact:

Theorem. Suppose that $G$ is a finite group such that $n_p(G)\gt 1$, and let $S$ and $T$ be Sylow $p$-subgroup of $G$ such that the order $|S\cap T|$ is as large as possible. Then $n_p(G)\equiv 1\pmod{|S\colon S\cap T|}$.

To see how your special case would follow from this, note that the condition $n_p(G)\not\equiv 1\pmod{p^2}$ means that $n_p(G)\gt 1$, and under the given assumption, it means that this “largest possible integersection” cannot have index $p^2$ or greater in $S$, and so you will be able to find a pair where $S\cap T$ has index $p$ in $S$ (and hence in $T$).

I won’t give you the entire argument, but instead just get your started. Note that if you can find two such Sylow $p$-subgroups $P$ and $Q$ as in your statement, and you have a favorite Sylow $p$-subgroups $T$, then by a suitable conjugation you find an example involving $T$: we know there is a $g\in G$ such that $gPg^{-1}=T$, and then $T$ and $gQg^{-1}$ will have the desired property.

That means that you can start with an arbitrary Sylow $p$-group $P$ and be sure that you can find a suitable $Q$ to match that one. So pick your favorite Sylow $p$-subgroup $P$.

Now let $P$ act on the set of all Sylow $p$-subgroups by conjugation. Note that one orbit is just $P$ itself; and that $P$ cannot fix any other Sylow $p$-subgroup (that is, you cannot have $P\subseteq N_G(Q)$ for a Sylow $p$-subgroup $Q\neq P$); if you don’t know that, then prove it. Use this fact to deduce something about the size of all other orbits, and use the fact that $n_p\not\equiv 1\pmod{p^2}$ to deduce that there must be an order of length $p$, which will give you the desired intersection.