Find the sum of all positive integers n which satisfy this condition

The following inequality are true for $n>15$ : $\large {(n+4)^2<n^2+9n+1<(n+5)^2}$ Actually, $(n+4)^2 = n^2+8n+16$ have the same degree of $n^2+9n+1$ but a smaller coefficient for $n$ , so after a precise value of $n$ (which is 15), $n^2+9n+1$ is a greater than $n^2+8n+16$ . So, for $n>15$ , $n^2+9n+1$ is a number between two squares and it cannot be a square. Now, we need to check all integer smaller than 15. The only solution is $\boxed{15}$ .

We require positive integers $n,m$ such that

$n^{2} + 9n + 1 = m^{2} \Longrightarrow (n + \frac{9}{2})^{2} - \frac{81}{4} + 1 = m^{2} \Longrightarrow (2n + 9)^{2} - 81 + 4 = 4m^{2} \Longrightarrow$

$(2n + 9)^{2} - (2m)^{2} = 77 \Longrightarrow ((2n + 9) + 2m)((2n + 9) - 2m) = 77$ .

Now $2n + 9 + 2m \gt 0$ , so to get a positive product we must also have that $2n + 9 - 2m \gt 0$ .

Then since $2n + 9 + 2m \gt 2n + 9 - 2m \gt 0$ , and since $77$ can be factored as a product of pairs of positive integers in only two ways, namely $77 \times 1$ and $11 \times 7$ , we must either have

• $2n + 9 + 2m = 77$ and $2n + 9 - 2m = 1 \Longrightarrow 2 \times (2n + 9) = 78 \Longrightarrow n = 15$ , or

• $2n + 9 + 2m = 11$ and $2n + 9 - 2m = 7 \Longrightarrow 2 \times (2n + 9) = 18 \Longrightarrow n = 0$ .

As the only positive integer solution is $15$ , the sum of all positive integer solutions is then $\boxed{15}$ .