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Let $x,y,z$ be positive integers such that $\text {HCF}(x,y,z)=1$ and $x^2+y^2=2z^2$. Prove the following

$1.$ $3$ divides $x+y$ or $3$ divides $x-y$

$2.$ $5$ divides $z(x^2-y^2)$

From $x^2+y^2=2z^2$, it is clear that $x$ and $y$ will be both odd or both even.

If both are even then $z$ will be odd so that $\text {HCF}(x,y,z)=1$ holds true but I think $x^2+y^2=2z^2$ could give more information which I am not able to observe and hence not able to prove the given statements. Could someone give some hint to proceed.

Maadhav
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Mathematics
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    https://math.stackexchange.com/questions/1767109/x2y2-2z2-positive-integer-solutions/1767719#1767719 – individ Dec 06 '17 at 15:59

1 Answers1

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b) By Fermat little theorem we have if $5\nmid a$ then $a^4 \equiv_5 1$ thus $a^2 \equiv_5 \pm1$.

If $5|z$ we are done, if $5\nmid 5$ then $2z^2\equiv_5 \pm 2$ so $x^2 \equiv_5 y^2 \equiv_5 1$ or $x^2 \equiv_5 y^2 \equiv_5 -1$ thus $5\mid x^2-y^2$. And we are done.

a) The same procedure can be done with mod $3$ of $3\nmid z$. If $3\mid z$ then $3\mid x^2+y^2$ then $3\mid x$ and $3\mid y$ and we are done.

nonuser
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