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
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
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 $\;n=1\;$:
$$x^2-2x-99=0\iff (x-11)(x+9)=0\;,\;\;\text{and etc.}$$