Re: Integer iff prime
> If p,q>3 are prime then p^2=1(mod 3) and q^2=1(mod 3), henceThat's what I meant to say.
> p^2+q^2-9=2(mod 3) is NEVER a square.
And p^2+2^2-9 = p^2-5 > (p-1)^2 for 2p-1>5 <=> p>3
so this can't be a square, either.
Which only leaves the trivial case q=3 => p^2 = p^2.
The formulation is a quite obscure way to say something which has not much to do with primes.
>Generalizing to the n-th degree, with a touch of Serendipity,
> When in doubt, Generalize.
> p^2 + q^2 - b^2 = k^2 has infinite solutions for b prime, b*p>3 p prime and no solutions for b a multiple of 4 or 6, maybe finite solutions b = p*3 or p*2, p prime.
> El Cino
1. p^n + q^n - b^n = k^2 has no integer solutions b, k for p, q prime, n odd > 3.
For p,q not both prime, there are many solutions.
p^n + q^n - b^n = k^m odd n > 3 has solutions but they are trivial, p, q = b.
Maybe someone can prove 1.
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