TITLE: Numbers of the form $(p_{1}^{\alpha_{_{_1}}})^{2}+(p_{2}^{\alpha_{_{_2}}})^{2}+\cdots+(p_{n}^{\alpha_{_{_n}}})^{2}=(p_{m}^{\alpha_{_{_m}}})^{2}$
QUESTION [9 upvotes]: I'm looking for numbers of the form
$$(p_{1}^{\alpha_{_{_1}}})^{2}+(p_{2}^{\alpha_{_{_2}}})^{2}+\cdots+(p_{n}^{\alpha_{_{_n}}})^{2}=(p_{m}^{\alpha_{_{_m}}})^{2}$$
where $p_{i}$ are prime numbers, $p_{i}\ne p_{j}$ and $\alpha_{_{k}}\in\mathbb{N}$
The first exemple is
$$(2^2)^2+(3^1)^2=(5^1)^2$$
I did a quick look at pythagorean triplets but could't find any.
So, I wonder if this is the only exemple or if there are finitely more or infinitely many.
What is known about this?

REPLY [2 votes]: As Gerry Myerson points out, there are no additional solutions with $n = 2$.
Since any odd term is congruent to 1 mod 8 and an even term is 0 or 4 mod 8, there can be no solutions for $3 \le n \le 5$.  (If the RHS is even, it is $> 4$ and thus 0 mod 8.)
Considerations mod 3 and mod 8 show that a solution with $n = 6$ is feasible only if the RHS is power of 3 and one of the terms in the LHS is 4.  A little playing around easily finds solutions such as
$$
2^2 + 7^2 + 17^2 + 23^2 + 47^2 + 59^2 = (3^4)^2
$$
and
$$
2^2 + 7^2 + 13^2 + 53^2 + 137^2 + 193^2 = (3^5)^2.
$$