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integer values of $n$ for which $x^2-(n+1)x+n = 100$ has integer roots

Attempts : assume $\alpha,\beta$ are the roots of given equation

so $\alpha+\beta = (n+1)$ and $\alpha \beta = n-100$

wan,t be able to go further, could some help me

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    $\begingroup$ Write $n-100=m$ and discuss the factors of $m$. $\endgroup$ – Yves Daoust Feb 14 '17 at 12:56
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Substract your second equation from the first one:

$$\alpha+\beta-\alpha\beta=101\iff(\alpha-1)(\beta-1)=-100$$

and as we have $\;\alpha,\,\beta\in\Bbb Z\;$ , you only need to check possibilities with the divisors of $\;100\;$

For example, $\;\alpha-1=-10\,,\,\beta-1=10\iff\alpha=-9\;,\;\;\beta=11\;$ , and we get $\;n=1\;$:

$$x^2-2x-99=0\iff (x-11)(x+9)=0\;,\;\;\text{and etc.}$$

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    $\begingroup$ Wow nice. I'm not sure I would have seen that. $+1$! $\endgroup$ – Cameron Williams Feb 14 '17 at 13:52

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