0
$\begingroup$

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 $k > 1$

  • 0
    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.2012-09-27
  • 0
    and if k is even?2012-09-27
  • 0
    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.2012-09-27
  • 0
    see edit please2012-09-27
  • 0
    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?)2012-09-27
  • 1
    @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 $a^k + b^k$"?2012-09-27
  • 0
    @Steve to the power of k/2, but i see both your examples work so back to the drawing board2012-09-27
  • 0
    im trying to prove that $$x^k + px = y^k$$ has only one solution2012-09-27
  • 0
    where p is prime and k>12012-09-27
  • 0
    There are certainly more than 1: $13=2^2+3^2$ and $41=4^2+5^2$.2012-09-27
  • 0
    @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.2012-09-27
  • 0
    http://math.stackexchange.com/questions/202247/number-of-solutions-to-equation2012-09-27

1 Answers 1

0

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.