Find the sum

There is a more general idea for solving these kinds of problems.

Let's say we have an equation of the form

$n^2+an+b = k^2$

with $a$ , $b$ , $n>0$ , $k \geq 0$ being integers. Rearranging the terms and examining the discriminant of the quadratic equation in terms of $n$ we end up with the following:

$n^2+an+b - k^2 = 0$

$D=a^2-4(b-k^2) = 4k^2+a^2-4b$

Since one of the solutions of the quadratic equation is integer, the only possibility is that the value of the discriminant must be a square of an integer. Hence for some $l$ nonnegative integer

$4k^2+a^2-4b = l^2$

$a^2-4b = l^2-4k^2$

$a^2-4b = (l-2k)(l+2k)$

The value of $a^2-4b$ is known, so upon factoring into prime components, through the investigation of several cases we should be able to calculate $l$ and $k$ , thus determine the possible values of $n$ .

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In the case of your example, we obtain the equation

$7 = (2k-l)(2k+l)$ .

Since 7 is prime and $l$ is nonnegative, we can only get in one hand $2k-l=1$ and $2k+l=7$ , or in the other hand $2k-l=-7$ and $2k+l=-1$ which yield $l=3$ and $k= \pm 2$ . Putting in the value of $k^2$ into the original equation we obtain the quadratic equation $n^2+19n+88=0$ with both roots being negative. If we allow the case $n \leq 0$ we obtain two solutions, namely -8 and -11. However, upon rejecting this possibility it follows that your example has no solutions over the positive integers.