# 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.

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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

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 $$2^2 + 7^2 + 13^2 + 53^2 + 137^2 + 193^2 = (3^5)^2.$$

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This is very similar to the Waring-Goldbach problem, which there are quite a few papers on (primarily when the $\alpha_i$'s are all equal but there are some results on mixed exponents), which asks how large $n$ must be in order to guarantee solutions to

$$p_1^\alpha + \cdots + p_n^\alpha = N,$$

for all sufficiently large $N$. In other words, there's nothing special about the RHS being a prime power: once you have a large enough (finite) number of terms, you can generate any number by summing prime powers together.

If we relax your requirement that the primes be distinct, then there are many solutions when $n$ is sufficiently large compared to $\alpha$. This web page suggests there are asymptotics available for $n \gg \alpha^2 \log \alpha$, presumably via the circle method. If so, then it's highly likely one could incorporate your restriction that the primes are all distinct: the asymptotics for the number of solutions where certain primes are equal will be much smaller when $N$ is very large compared to $n$.

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