Problems & Puzzles: Puzzles

Puzzle 493. p^p = x*a^2 + y*b^2 Michael Hiebl sent the following puzzle: There is always solution for p^p = x*a^2 + y*b^2 p = prime>2 x + y = p (a, b, x, y)=integers gcd(a,b)=1 a<>b (a, b)>0 Q. Is this true?

---

***

On October 29, 2026, Bas van Dorst launched a massive attack on 61 puzzles (64 results), including this one. Please see his contribution to this puzzle on the page https://www.primepuzzles.net/puzzles/puzz_1288.htm

***