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If we have a prime number $p$ and two natural numbers $a$ and $b$ such that $p^a=b^4+4$, then how many such ordered triplets $(p,a,b)$ exist?

What should be the strategy to solve this one? The only I can see is $(5,1,1)$, is this the only one? if yes, how could we prove that?

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oeis.org/A057781 –  msh210 Dec 23 '11 at 19:33
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The following perhaps surprising factorization may be useful: $b^4+4=(b^2-2b+2)(b^2+2b+2)$. –  André Nicolas Dec 23 '11 at 19:35
    
@J. M.:Believe it or not this was a quantitative aptitude problem. ;) –  Quixotic Dec 24 '11 at 7:05
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2 Answers

up vote 5 down vote accepted

This is an old contest problem, I wish I could remember where I first saw it. Anyway, André's comment that $$b^4+4=(b^2-2b+2)(b^2+2b+2)$$ is the key to a solution.

Looking modulo $16$, we see that $b^4+4$ cannot be a power of $2$. For $b>1 $, both factors will be strictly greater then $1$, so that if $p^k|(b^4+4)$ then $p$ must divide both $b^2-2b+2$, and $b^2+2b+2$. Since $\gcd(b^2-2b+2,b^2+2b+2)$ must divide $4b$, and $p$ divides both terms, we see that $p|b$. This then implies that $p$ divides $4$ which is impossible.

If $b=1$, then we get the one special case.

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You are saying $b^4 +4 \equiv 4 \text{ or }5 \mod{16}$, but $4$ is a power of $2$. –  Henry Dec 23 '11 at 22:15
    
@Henry I wasn't counting $b=0$ as a natural number. –  Eric Naslund Dec 23 '11 at 23:10
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It depends on your definition of natural number, but $(2,2,0)$ may be the only other possibility

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Perhaps this is more suited as a comment? –  Quixotic Dec 24 '11 at 7:13
    
@MaX: Perhaps, but it is an answer –  Henry Dec 24 '11 at 11:51
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