$q=2p+1$ and $p\equiv 3 \mod 4$. Show that $q|2^p-1$.

Question

Let $p,q$ prime numbers such that $q=2p+1$ and $p\equiv 3 \mod 4$. Show that $q|2^p-1$.

Approach: We have $q\equiv 3 \mod 4$, $q\equiv 1 \mod p$. We also have $$\left(\frac{p}{q}\right) = -1$$. Any hints? Thanks ind advance!

Answer 1

$2^p=2^{\frac{q-1}{2}}\equiv \big(\frac{2}{q}\big)\bmod q$ (Already noted in comments)

Now $q=2(4k+3)+1$

Hence $q$ leaves a remainder $7$ when divided by $p$, hence $\big(\frac{2}{q}\big)=1$.

(See this http://mathonline.wikidot.com/legendre-symbol-rules-for-1-p-and-2-p)

Hence $q|2^p-1$.