Use the basics

Number Theory Level 4

Find the sum of all positive integers $d$ such that

$d\mid(n^2+1)$

$d\mid[(n+1)^2+1]$

Details & Assumptions

$\bullet$ $n$ is a non-negative integer

$\bullet$ $a \mid b$ means " $a$ divides $b$ " i.e. $b$ is divisible by $a$ .

1 solution

Khang Nguyen Thanh
Aug 6, 2015

Since $d\mid(n^2+1)$ and $d\mid[(n+1)^2+1]$ , we get $d|[(n+1)^2+1-(n^2+1)$ or $d\mid 2n+1$ .

Thus, $d\mid (2n+1)(2n-1)$ or $d\mid(4n^2-1)$ .

On the other hand, $d\mid4(n^2+1)$ , implies $d\mid[4(n^2+1)-(4n^2-1)$ or $d\mid5$ .

Since $d$ is a positive integer, so $d\in\{1; 5\}$ .

If $n=1$ , then $d=1$ .

If $n=2$ , then $d=5$ .

So both of above value of $d$ are satisfied.

Hence, the answer is $1+5=\boxed{6}$ .