# Numbers of the form $(p_{1}^{\alpha_{_{_1}}})^{2}+(p_{2}^{\alpha_{_{_2}}})^{2}+\cdots+(p_{n}^{\alpha_{_{_n}}})^{2}=(p_{m}^{\alpha_{_{_m}}})^{2}$

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.

You won't find any more looking at Pythagorean triplets. You'd need $2mn$ a power of 2, and $m^2-n^2$ (and $m^2+n^2$) prime. But $m^2-n^2=(m+n)(m-n)$ can only be prime if $m=n+1$, and then $2mn$ can't be a power of 2 except when $n=1$. – Gerry Myerson Nov 5 '12 at 11:40
@GerryMyerson You make a fair point, but that assumes $\alpha_1 = \alpha_2 = 1$. For a more general "you can't look at Pythagorean triplets", you need to assume that $m^2 + n^2$ and $m^2 - n^2$ is some power of some prime. – Arthur Nov 5 '12 at 11:57
@Arthur, you're right --- but $m^2-n^2$ can't be a power of an odd prime unless $m=n+1$, or $m$ and $n$ are both divisible by that prime, and again $2mn$ can't be a power of 2. – Gerry Myerson Nov 5 '12 at 12:15