# About hypotenuse of pythagorean triple [duplicate]

The hypotenuse of any Pythagorean triple seems not to be divided by some primes such as $3, 7$. What are the others? Are there infinitely many such primes?

In other words, I am looking for number $c$ such that $a^2 + b^2 = kc$ have no integer solutions for any positive integer $k$ and $a,b < c$.

## marked as duplicate by lulu, Matthew Towers, Andrew D. Hwang, Gerry Myerson number-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 3 '17 at 11:44

• Yes, these are the primes of the form $4n+3$ and there are infinitely many of them. – Ivan Neretin Feb 3 '17 at 11:21
• There are an infinite number of solutions of pythagorean triples. Sir Wiles proved that for the equation $x^{n} + y^{n} = z^{n}$ has no solutions for n larger than two. – Cppg Feb 4 '17 at 7:49

Suppose $a$ and $b$ are integers. I that case euclid formulas hold: \begin{align} a &= m^2-n^2 \\ b &= 2mn \\ c &= m^2+n^2 \end{align} where $m,n$ are positive integers.

Now use the following Fermat theorem (have a look at Arturo's answer):

Theorem (Fermat) Let $n$ be a positive integer, and write $n$ in the form $$n = 2^{\alpha}\prod_{p\equiv 1\pmod{4}} p^{\beta} \prod_{q\equiv 3 \pmod{4}} q^{\gamma}.$$ with $p$ and $q$ primes. Then $n$ can be expressed as a sum of two squares if and only if all the exponents $\gamma$ are even.

Since you are looking for only primes then it becomes $n = p$ such that $p \equiv 3 \pmod 4$.

Look for primes that can be written as $n = 4k + 3$ with $n$ integer.