7. $p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}+p_{5}^{2}=p_{6}^{2}$ has $\qquad$ groups of positive prime solutions $\left(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}\right)$.
7.5.

Obviously, $p_{6} \neq 2$.
Since $p^{2} \equiv 1$ or $4(\bmod 8)$ (where $p$ is a prime), and $p^{2} \equiv 4(\bmod 8)$ if and only if $p=2$, it follows that among $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, and $p_{5}$, there must be 4 twos (let's assume they are $p_{1}$, $p_{2}$, $p_{3}$, and $p_{4}$), and the other one is an odd prime, and $p_{6}$ is also an odd prime.
Thus, $16=p_{6}^{2}-p_{5}^{2}=\left(p_{6}+p_{5}\right)\left(p_{6}-p_{5}\right)$.
Clearly, the only solution is $p_{5}=3, p_{6}=5$.
Therefore, the number of all prime solutions is $\mathrm{C}_{5}^{4}=5$.