# 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 $$2 \times 10^9$$

• What is the source for this claim? Dec 20, 2019 at 5:27
• @EricWofsey It my own problem, no other soruce that I am aware of Dec 20, 2019 at 5:33
• You should not call it a "claim" if you do not know it is true! Dec 20, 2019 at 5:34
• @EricWofsey Alright, conjecture then Dec 20, 2019 at 5:35

## 1 Answer

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.