## Re: Integer iff prime

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• ... That s what I meant to say. 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
Message 1 of 6 , Apr 25, 2009
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> If p,q>3 are prime then p^2=1(mod 3) and q^2=1(mod 3), hence
> p^2+q^2-9=2(mod 3) is NEVER a square.

That's what I meant to say.

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.

Maximilian
• ... Generalizing to the n-th degree, with a touch of Serendipity, 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
Message 2 of 6 , Apr 25, 2009
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>
> 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
>

Generalizing to the n-th degree, with a touch of Serendipity,

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.

Cino

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