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(B) Given $x^2+y^2=2z^2$ with $HCF(x, y, z)=1$.Check $I$: Let $x=1, y=7, z=5$. Here $1^2+7^2=50=2(5^2)$. $HCF(1, 7, 5)=1$. $4$ does not divide $1$ or

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$II$ and $III$ only

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(B) Given $x^2+y^2=2z^2$ with $HCF(x, y, z)=1$.Check $I$: Let $x=1, y=7, z=5$. Here $1^2+7^2=50=2(5^2)$. $HCF(1, 7, 5)=1$. $4$ does not divide $1$ or $7$. Thus,$I$ is false.Check $II$: $x^2+y^2=2z^2 \implies x^2+y^2 \equiv 2z^2 \pmod 3$. Squares modulo $3$ are $0, 1$. If $z^2 \equiv 0$,then $x^2+y^2 \equiv 0 \implies x, y$ are multiples of $3$,contradicting $HCF=1$. If $z^2 \equiv 1$,then $x^2+y^2 \equiv 2 \implies x^2 \equiv 1, y^2 \equiv 1$. Then $x \equiv \pm 1, y \equiv \pm 1$. Thus $x+y \equiv 0$ or $x-y \equiv 0 \pmod 3$. $II$ is true.Check $III$: $x^2+y^2=2z^2 \implies x^2-z^2 = z^2-y^2$. Modulo $5$,squares are $0, 1, 4$. If $z^2 \equiv 0$,then $x^2+y^2 \equiv 0 \implies x, y \equiv 0$,contradiction. If $z^2 \equiv 1$,$x^2+y^2 \equiv 2$. Possible pairs $(x^2, y^2)$ are $(1, 1)$. Then $x^2-y^2 \equiv 0 \pmod 5$. If $z^2 \equiv 4$,$x^2+y^2 \equiv 8 \equiv 3$. Possible pairs $(x^2, y^2)$ are $(4, 4)$. Then $x^2-y^2 \equiv 0 \pmod 5$. In both cases,$5$ divides $x^2-y^2$,so $5$ divides $z(x^2-y^2)$. $III$ is true.

Let $x, y, z$ be positive integers such that $HCF(x, y, z)=1$ and $x^2+y^2=2z^2$. Which of the following statements are true?
$I$. $4$ divides $x$ or $4$ divides $y$.
$II$. $3$ divides $x+y$ or $3$ divides $x-y$.
$III$. $5$ divides $z(x^2-y^2)$.

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Let $x, y, z$ be positive integers such that $HCF(x, y, z)=1$ and $x^2+y^2=2z^2$. Which of the following statements are true?$I$. $4$ divides $x$ or $4$ divides $y$.$II$. $3$ divides $x+y$ or $3$ divides $x-y$.$III$. $5$ divides $z(x^2-y^2)$.