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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 $a,b < c$.

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marked as duplicate by lulu, Matthew Towers, Andrew D. Hwang, Gerry Myerson number-theory Feb 3 '17 at 11:44

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  • $\begingroup$ Yes, these are the primes of the form $4n+3$ and there are infinitely many of them. $\endgroup$ – Ivan Neretin Feb 3 '17 at 11:21
  • $\begingroup$ 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. $\endgroup$ – Cppg Feb 4 '17 at 7:49
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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.

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