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How many primes are there of the form $a^{k/2} + b^{k/2}$ exist for $a$ and $b$ (positive integer solutions).

I am hoping there is only one.

EDIT $k > 1$

Michael Hardy
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fosho
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  • If $k$ is odd and greater than $1$, there is none since $a+b$ divides $a^k + b^k$. And every prime of the form $4m+1$ can be expressed as a sum of two squares. –  Sep 27 '12 at 21:28
  • and if k is even? – fosho Sep 27 '12 at 21:28
  • If $a=2$ and $b=1$ there are known to be multiple solutions, the so-called Fermat primes. If $a=2$ and $b=3$ then $k=1$, $k=2$ and $k=4$ all give solutions. – Steven Stadnicki Sep 27 '12 at 21:30
  • see edit please – fosho Sep 27 '12 at 21:31
  • What is meant by the initial equation (since right now the syntax is a bit messed-up)? Can you give a good specific example? (And explain why $2^2+3^2$ and $2^4+3^4$ both being prime doesn't contradict your hopes?) – Steven Stadnicki Sep 27 '12 at 21:33
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    @fosho Is your question "Given $a$ and $b$, how many primes are of the form $a^k + b^k$?" or "Given $k$, how many primes are of the form $a^k + b^k$?" or is it just "How many primes are of the form $a^k + b^k$"? –  Sep 27 '12 at 21:34
  • @Steve to the power of k/2, but i see both your examples work so back to the drawing board – fosho Sep 27 '12 at 21:36
  • im trying to prove that $$x^k + px = y^k$$ has only one solution – fosho Sep 27 '12 at 21:37
  • where p is prime and k>1 – fosho Sep 27 '12 at 21:37
  • There are certainly more than 1: $13=2^2+3^2$ and $41=4^2+5^2$. – 2'5 9'2 Sep 27 '12 at 21:39
  • @fosho you might as well write $a^k+b^k$, since non-integer exponents can't produce rational results. If you're trying to take into account that any $k\gt 1$ must be even (as Marvis notes) then you presumably want $a^{2k}+b^{2k}$ - but I wouldn't even bother with that; instead I would just write $a^k+b^k$ and note the implicit restriction on $k$ as an aside. – Steven Stadnicki Sep 27 '12 at 21:39
  • http://math.stackexchange.com/questions/202247/number-of-solutions-to-equation – fosho Sep 27 '12 at 21:43

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Infinitely many. In fact, every prime $p \equiv 1 \pmod 4$ can be written as the sum of two squares; a result attributed to Fermat. And there are infinitely many such primes, according to Dirichlet's Theorem.

Douglas S. Stones
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