<s>[INST] If two side of a Pythagorean triangle are primes then the third side has at least $4$ distinct prime factors. 
Conjecture: If two side of a Pythagorean triangle are primes $> 19$ then the third side has at least $4$ distinct prime factors.

E.g. $421^2 = 420^2 + 29^2$. Here $29$ and $421$ are both primes and the third side $420$ has four distinct prime factors $2,3,5,7$.
I can show that the third side is divisible by $60$ which contributes the three prime factors $2,3$ and $5$. How do we show that there is always a fourth prime?
Update: Verified this for primes upto $2 \times 10^9$
 [/INST] If $a^2+b^2=c^2$ where $a,b,c$ are relatively prime positive integers, with $b$ even, then there are integers $x,y>0$ such that $a=x^2-y^2$, $b=2xy$, and $c=x^2+y^2$.  In your case, $a$ and $c$ must be primes $>19$.  In particular, for $a=(x-y)(x+y)$ to be prime, we must have $x-y=1$ so we have $a=2y+1$, $b=2y(y+1)$, and $c=2y^2+2y+1$.  The claim is then that if $a$ and $c$ are primes greater than $19$ then $b$ has at least $4$ distinct prime factors.
Simple mod $3$ and mod $5$ considerations show that $b$ must be divisible by $3$ and $5$.  If $b$ has at most $3$ distinct prime factors, then, both $y$ and $y+1$ have no prime factors greater than $5$.  It can be shown that there are only ten positive integers $y$ such that both $y$ and $y+1$ have no prime factors greater than $5$, namely $y=1,2,3,4,5,8,9,15,24,80$.  Testing all of these values of $y$, none yield values of $a$ and $c$ that are both primes greater than $19$, so there are no examples where $b$ has only three distinct prime factors.
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