7. Given prime numbers $p$ and $q$ satisfy $q^{5}-2 p^{2}=1$. Then $p+q=$ $\qquad$

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7. 14 .
$$
\begin{aligned}
\text { Given } & q^{5}-2 p^{2}=1 \\
\Rightarrow & 2 p^{2}=q^{5}-1 \\
& =(q-1)\left(q^{4}+q^{3}+q^{2}+q+1\right) .
\end{aligned}
$$

It is easy to see that \( q^{4}+q^{3}+q^{2}+q+1 \) is an odd number greater than 1, and \( p^{2} \) is also an odd number greater than 1.
$$
\begin{array}{l}
\text { Then } q-1=2 \Rightarrow q=3 \\
\Rightarrow 2 p^{2}=3^{5}-1=242 \Rightarrow p=11 \\
\Rightarrow p+q=14 .
\end{array}
$$