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"Three positive integers $x$, $y$ and $z$ are called a Pythagorean triple if $x^2 + y^2 = z^2$. A Pythagorean triple is called primitive if the only positive integer that is a factor of all three integers $x$, $y$ and $z$ is $1$. Explain why, if $x$, $y$ and $z$ is a primitive Pythagorean triple, either $x$ or $y$ must be odd." - From book

An exercise from my proof book that I'm confused about. I'm a beginner any help would be appreciated

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  • $\begingroup$ Also $x,y$ cannot both be odd:... If $x,y$ are both odd then the remainders of $x^2$ and $y^2$ when divided by $4$ will each be $1$ , so the remainder of $x^2+y^2$ when divided by $4$ will be $2,$ so $x^2+y^2$ cannot be $z^2$..... Because the remainder of $z^2$ when divided by $4$ must be $0$ or $1.$ $\endgroup$ – DanielWainfleet Sep 13 '17 at 17:03
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Suppose both $x$ and $y$ are even then $x^2$ and $y^2$ will be even and their sum $z^2=x^2+y^2$ will also be even. That implies that $z$ is also even. So all of $x$, $y$, and $z$ are divisible by $2$. Hence, their gcd is not $1$. A contradiction.

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In fact $x$ and $y$ must be coprime. If they have any common factor $d$ then $d^2$ divides $x^2$ and $y^2$ and thus also divides $z^2$ and so $d$ divides $z$, contradicting the definition of a primitive triple. The particular case of any even $d$, including $d=2$, shows that both $x$ and $y$ cannot be even.

The reason it is worth thinking about primitive triples is that each such $(x,y,z)$ generates a whole family of triples $(kx,ky,kz), k\in \Bbb N$. Of course when $k$ is even, the whole resulting triple is even.

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