If $a,b,c$ are positive integers and $a^2+b^2=c^2$ and $a$ is a prime, what can we conclude about primeness of b and c? Let $a,b,c$ be positive integers and they satisfy $a^2+b^2=c^2$, and if $a$ is prime, can we conclude whether $b$ and $c$, are both prime, composite or neither? If yes, why, if not why not?
I can conclude that $b$ and $c$ have to be one odd and the other one even using $a^2 = c^2-b^2=(c-b)(c+b)$. But I couldn't conclude anything about their primeness. Can anyone show me some ideas or its reasoning and maybe any useful and related theorems?
Thanks very much!
 A: Pythagorean triples are always of the form
$$\begin{align}
a&=(u^2-v^2)d\\
b&=2uvd\\
c&=(u^2+v^2)d
\end{align}$$
where $\gcd(u,v)=1$ (and the expressions for $a$ and $b$ can be swapped).  If $a$ is a prime, we quickly see that $d$ must be $1$ and $u^2-v^2=(u-v)(u+v)=1(2v+1)$, the upshot of which is
$$\begin{align}
b&=(a^2-1)/2\\
c&=(a^2+1)/2\\
\end{align}$$
The $b$ can never be prime, but $c$ can.  The first few correspond to $a=3,5,11$, and $19$ (with corresponding $c=5,13,61$, and $181$).  More are given in the OEIS at A048161.
A: Since $a$ is a prime and as $a$ is not equal to $2$, $a$ is odd.
Since $c^2$ leaves residue $1,0$ on division by $4$, hence $b$ is even and is composite.
$c$ can either be prime or composite with examples $3,4,5$ and $7, 24, 25$

I believe if there were some condition for $a$ and $b$ for which $c$ should be prime, then we could have found out a prime number generator!
A: $c$ is prime and $b$ isn't :
$$3^2+4^2=5^2$$ 
both are composite :
$$7^2+24^2=25^2$$
However, $b$ can never be a prime. Proof : 
$a=2$ or $b=2$ is impossible (first pythagorean triplet is $3,4,5$)
if both $a$ and $b$ are odd prime numbers, then $a^2\equiv b^2\equiv 1 \mod 4$ (this is true for all odd numbers), and therefore $c^2\equiv 2\mod 4$ (impossible)
A: $5^2+12^2=13^2$ and $7^2+24^2=25^2$, so I don't think there is anything interesting about primality of $b$ and $c$...
What you may say is that, except for trivial cases, $c+b\neq c-b$, hence we must have $c-b=1$ and $c+b=2b+1=a^2$, so $b=\frac{a^2-1}2$ and $c=\frac{a^2+1}2$. Hence, you find that exists exactly one such triple $(a,b,c)$ for every odd prime number.
