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Suppose $a$, $b$, and $c$ are three distinct integers which share no common divisor. Note that this implies that $a$, $b$, and $c$ are pairwise coprime. If $a$,$b$, and $c$ are a primitive Pythagorean triple satisfying the equation $a^2 + b^2 = c^2$ , is it then always true that

$$\gcd ( (a+b) , (a+b-c) ) = 1?$$

Can it be proven that $\gcd ( (a+b) , (a+b-c) ) = 1$ invariably ?

Can you find a counterexample where $\gcd ( (a+b) , (a+b-c) ) \neq 1$?

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We have that $$ \gcd(a+b, a+b-c) = \gcd(a+b, -c) = \gcd(a+b, c). $$ Now suppose that $p$ is a prime dividing $\gcd(a+b, c)$. Because $a^2 + b^2 = c^2$, we have that $p^2$ also divides $(a+b)^2 - (a^2 + b^2) = 2ab$. In particular $p$ divides either $a$ or $b$ (even if $p=2$). But this contradicts our assumption.

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    $\begingroup$ You forgot to (explicitly) eliminate the possibility $\,p = 2.\ \ $ $\endgroup$ Aug 23 '19 at 13:26
  • $\begingroup$ @BillDubuque, I don't treat it as a special case; note that $p^2$ divides $2ab$, so even if $p=2$, that still leaves a factor of 2 for $ab$. $\endgroup$ Aug 23 '19 at 13:32
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    $\begingroup$ Yes but the peculiarity for $\,p = 2\,$ deserves explicit mention (else the reader can't know whether or not there is actually a gap in the argument). I'd write "(even if $p = 2$)" $\endgroup$ Aug 23 '19 at 13:45
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Hint $\ (a\!+\!b, a\!+\!b\!-\!c)= (a\!+\!b,c)\mid (a\!+\!b,\!\!\!\overbrace{c^2}^{a^2\,+\,b^2}\!\!\!)\mid 2(a^2,b^2)\!=\!2,\,$ but $c$ is odd in a PPT

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    $\begingroup$ See also here for a proof that $\,(a\!+\!b,\,a^2\!+\!b^2)\mid 2(a,b)^2\ $ and more. $\endgroup$ Aug 23 '19 at 14:27

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