7. Proof: Since $p \equiv 1(\bmod 4)$, -1 must be a quadratic residue modulo $p$, i.e., there exists an integer $x_{0}$ such that
$$x_{0}^{2} \equiv -1(\bmod p)$$

On the other hand, from $p=4 q+1$ we also have
$$4 q \equiv -1(\bmod p)$$

Therefore,
$$x_{0}^{2} \equiv 4 q(\bmod p)$$

Since $2 \nmid p$, there must be a $y_{0}$ such that $2 y_{0} \equiv 1(\bmod p)$, thus
$$\left(x_{0} y_{0}\right)^{2} \equiv q\left(2 y_{0}\right)^{2} \equiv q(\bmod p)$$

This is exactly what we need to prove.