A number theory problem by Anubhav Tyagi

just do polynomial division .

then $n^{2} + 11n + 2 = (n + 5)(n + 6) - 28.$

It follows that $-28$ should be divisible by $n + 5$

So, the answer is the number of factors of $-28$ , considering only positive solutions of n.

Transform into another form:

$\frac { { n }^{ 2 }+11n+2 }{ n+5 } =\frac { { (n+5) }^{ 2 }+(n+5)-28 }{ n+5 }$

So, $-28$ must be divisible by $n+5$ , thus $(n+5)\in \left\{ 7;14;28 \right\}$ , because $n$ is a positive integers. In conclusion, possible solutions for $n$ are $2;9;23$ .

Given polynomial $\frac { { n }^{ 2 }+11n+2 }{ n+5 }$

Substitute n+5=t . So, the polynomial becomes $\frac { { (t-5) }^{ 2 }+11(t-5)+2 }{ t }$ $=\frac { { t }^{ 2 }+t-28 }{ t }$ $=t+1-\frac { 28 }{ t }$

Now, t+1 is always an integer for all integers t . So, we have to ensure that $\frac { 28 }{ t }$ is also an integer. $\frac { 28 }{ t }$ is an integer for t=1,2,4,7,14,28 giving us values of n as n=-4,-3,-1,2,9,23 out of which only 3 are positive.