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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 $2 \times 10^9$

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

1 Answer 1

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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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