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Given that $a^2+b^2=c^2$ and $c$ is prime, is there a value for the shortest side of this right triangle after which the hypotenuse is never again prime? Or is there always a larger prime?

I ask because in the $b+1$ sequence, values for $c=b+1$ begin $5$, $13$, $25$, $41$, $61$, $85$, etc., and primes occur often. But do they occur perpetually? And if they do, is this true for all Pythagorean sequences? Including the $b+2$ sequence? When $c=b+2$ and the value for $a$ is divisible by $4$, the sequence of $c=b+2$ begins $5$, $17$, $37$, $65$, $101$, $145$, $197$, $257$, etc.

Doesn't this mean there is an infinitude of primes of the form $x^2+1$?

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Please don't use all-caps for emphasis (it's considered shouting). You can use *italics* (italics) or **bold** (bold) in the text (though not in the title). –  Arturo Magidin Apr 4 '12 at 15:27
    
@dcc: An interesting way to rephrase the conjecture that there are infinitely many primes of the form $x^2+1$! –  André Nicolas Apr 4 '12 at 15:33
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2 Answers

Every prime of the form $4k+1$ is the hypotenuse of some primitive pythagorean triple (and there are an infinite of those).

This is because we have $\{a,b\} = \{m^2-n^2, 2mn\}$ and $c = m^2 + n^2$ and Fermat proved that every prime which of the form $4k+1$ is the sum of two squares.

(Or look at it terms of Gaussian Integers).

For $c = m^2 + n^2$, what you are asking in the later part amounts to showing that $\min\{m,n\} = O(1)$ (See OEIS) and would be implying a weaker form of the conjecture that there is always a prime between two consecutive perfect squares. So I doubt there will be an easy proof of that, or if that is even true.

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btw, did I misunderstand your first question itself? –  Aryabhata Apr 4 '12 at 16:25
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If $c=b+1$, and $c$ is prime, let's write $p$ for $c$, and $p-1$ for $b$, so $$a^2+(p-1)^2=p^2$$ which is $p=(a^2+1)/2$. So in this case you are asking whether there are infinitely many primes of the form $(x^2+1)/2$. While the answer is certainly yes, and there are even good estimates for the number of such primes up to any given limit, no one has yet been able to give a proof.

Similarly, your $c=b+2$ reduces to $p=x^2+1$, and the same remarks apply.

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