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Let $(a, b, c)$ be a primitive Pythagorean triple. I know that $\gcd(a,b,c) = 1$. Is it always true that $\gcd(a,b) = \gcd(b,c) = \gcd(a,c) = 1$?

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The answer is yes, but just because $(a,b,c)$ is a primitive Pythagorean triple.

Assume for contradiction that $\gcd(a,b)>1$. Then $\exists p\in\mathbb P(p\mid a,b\Rightarrow p\mid a^2+b^2=c^2\Rightarrow p\mid c\Rightarrow \gcd(a,b,c)\ge p)$, contradiction.

If $a,b,c$ weren't a primitive Pythagorean triple, a counterexample would be $(a,b,c)=(p_1p_2,p_2p_3,p_3p_1)$, $p_1,p_2,p_3\in\mathbb P$.

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Suppose that $(a,b,c)$ is a Pythagorean triple and that $p$ is a prime number that divides two of these numbers (say $a$ and $b$). By the Pythagorean theorem $p|c^2$ too, and since $p$ is prime $p|c$.

So yes, it is true.

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  • $\begingroup$ Yes, but this answer is phrased to be a general statement about Pythagorean triples. In the particular case of primitive triples it answers the OP. $\endgroup$
    – Umberto P.
    Apr 13 '15 at 20:01

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