How many solutions are there to a^2 + b^2 = c^5 for integers a, b, c?

arithmetic/two.squares.odd.power New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

How many solutions are there to a^2 + b^2 = c^5 for integers a, b, c?
How about a^2 + b^2 = c^7?

Show solution

Infinitely many, even with a,b positive and relatively prime.
Start with x,y relatively prime of opposite parity, and obtain
a,b,c from a+bi=(x+yi)^5 [or (x+yi)^7], c=x^2+y^2 [Editor: "c=a^2+b^2"
in the original].  You may have to change the signs of a and/or b.
Example: x=1, y=2 [Editor: "a=1, b=2" in the original] yields
(1+2i)^5 = 41 - 38i  and  (1+2i)^7 [Editor: "(1+2i)67" in the
original] = 29 + 278i, whence 41^2 + 38^2 = 5^5  and
29^2 + 278^2 = 5^7.  The proof that this works is left as an exercise
(or puzzle, if you prefer).

--Noam D. Elkies (elkies@math.harvard.edu)
  Dept. of Mathematics, Harvard University