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$x,y>2$ and $\in\mathbb{N}$. Show,that if $x^2+y^2-1$ is divided by $x+y-1$, then $x+y-1$ isn't prime.

$$x^2+y^2-1=(x+y)^2-1-2xy=(x+y-1)(x+y+1)-2xy$$ Thefore, $2xy=k(x+y-1), k\in\mathbb{N}$. Some steps more: if $x+y-1$ is even, then it's nothing to talk about, so let $x+y-1$ is odd. Then $xy=k'(x+y-1), k'\in\mathbb{N}$, moreover, $x$ and $y$ are both even or both odd. And...I stuck.

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How can $(x+y-1)$ divide $2xy$ ? – OBDA Mar 15 '14 at 23:07
up vote 4 down vote accepted

Let $p$ be a prime. If $p$ divides a product, $p\mid abc$, then $p$ divides one of the factors, $p\mid a$ or $p\mid b$ or $p\mid c$ (Euclid's lemma).

Now if $x+y-1$ were prime ...

... it would divide $2,\; x$, or $y$. But it is greater than all of the three.

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then it divides $x$ or $y$, then $x$ or $y$ isn't prime...and i don't reach any contrаdiction. /: – Evgeny Egorov Mar 15 '14 at 23:08
Since $x,y \geqslant 2$, can you have $(x+y-1) \leqslant x$? – Daniel Fischer Mar 15 '14 at 23:10
No, I can't. It's greater than both $x$ and $y$... – Evgeny Egorov Mar 15 '14 at 23:13
I get it! Thanks! – Evgeny Egorov Mar 15 '14 at 23:15

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