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Find the number of all integer-sided isosceles obtuse-angled triangles with perimeter $2008$.

My efforts:

let sides be $x$, $x$, $y$ so $2x+y=2008$ , but by triangle inequality $y<2x$ also since it is obtuse-angled we can $y>x$.

these were my deductions, so how do I proceed further? any help is appreciated.

Thanks in advance!

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    $\begingroup$ You can continue with saying that $y<1004$ and if $2y$ is even $x$ is even too. $\endgroup$ – Deniz Tuna Yalçın Oct 9 '17 at 16:55
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To have it obtuse, you need $y \gt x \sqrt 2$. Also $y$ must be even so that you can divide $2008-y$ by $2$. What is the smallest $y$ that works? What is the largest?

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    $\begingroup$ isn't it easier to work out the smallest and largest $x$? $\endgroup$ – zwim Oct 9 '17 at 17:23

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