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?

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