How many perfect square?

Note that for positive integers $x$ we have that

$(2x^{2} + x)^{2} = 4x^{4} + 4x^{3} + x^{2} \lt 4x^{4} + 4x^{3} + 4x^{2} + 4x + 4$ .

Also, we have that

$(2x^{2} + x + 1)^{2} = 4x^{4} + 4x^{3} + 5x^{2} + 2x + 1 \gt 4x^{4} + 4x^{3} + 4x^{2} + 4x + 4$

whenever $x^{2} \gt 2x + 3 \Longrightarrow x^{2} - 2x - 3 \gt 0 \Longrightarrow (x + 1)(x - 3) \gt 0$ ,

which is the case for $x \gt 3$ , (and also $x \lt -1$ , but this is not relevant to this question).

So letting $S = x^{4} + x^{3} + x^{2} + x + 1$ and $m = 2x^{2} + x$ we have that, for $x \gt 3$ ,

$m^{2} \lt 4S \lt (m + 1)^{2} \Longrightarrow \left(\dfrac{m}{2}\right)^{2} \lt S \lt \left(\dfrac{m + 1}{2}\right)^{2}$ .

Now if $x$ is even, thus making $m = 2k$ even, then $k^{2} \lt S \lt (k + \frac{1}{2})^{2} \lt (k + 1)^{2}$ ,

and if $x$ is odd, thus making $m = 2k + 1$ odd, then $k^{2} \lt (k + \frac{1}{2})^{2} \lt S \lt (k + 1)^{2}$ .

Either way, for $x \gt 3$ we have that $S$ lies strictly between two perfect squares, and thus cannot be a perfect square itself. So we are left to simply check the cases $x = 1,2,3$ , which in turn yield values for $S$ of $5, 31$ and $121 = 11^{2}$ , the latter being the only perfect square. Thus as $x = 3$ is the only positive integer that makes $S$ a perfect square, the sum of all such values is $\boxed{3}$ .