sum of primes squared equals a prime squared

Question

Find all $p_1,\dots,p_8$ so that $$ p_1^2+\dots+p_7^2 = p_8^2 $$ What I have done so far:
Since an odd number $n$ has $n^2 \equiv 1$ (mod $4$) then the only two possibilities are if $2$ or $6$ of the primes on the left are $2$. When $6$ of them are $2$, then I found a solution when $p_7 = 5$ and $p_8=7$. I cannot figure out the other case though. I got to $$ p_3^2 + p_4^2 + p_5^2 + p_6^2 + p_7^2 = p_8^2-8 $$ but that's it. But I cannot find a way to find them all or prove none exist.

Answer 1

As user113102 hints, if $n$ is odd, then $n^2 \equiv 1 \bmod 8$. Thus, in the equation $$p_3^2 + p_4^2 + p_5^2 + p_6^2 + p_7^2 = p_8^2 - 8$$ where all $p_i$ are odd primes, the LHS is congruent to $5$ mod $8$ and the RHS is congruent to $1$, so there is no solution.

It's easy to see, therefore, that the solution $p_1 = \cdots = p_6 = 2, p_7 = 5, p_8 = 7$ is therefore the only solution to the general case, as larger values of $p_7$ and $p_8$ would give $p_8^2 - p_7^2 > 24$.