Square valued integer polynomial

Question

For what integer values of $n$ is the expression $n^{6}+n^{4}+1$ a square?

This is a square when $n=2$; when $n$ is odd, this expression is $3 \ mod(8)$ and so cannot be a square.

Answer 1

Like you noted, $n$ odd doesn't work.

For $n$ even, note that looking only at non-negative integers is sufficient. We have, for $n\ge 4$,

$$\left(n^3+\frac{n}{2}\right)^2=n^6+n^4+\frac{n^2}{4}> {\bf n^6+n^4+1}> n^6+n^4-2n^3+\frac{n^2}{4}-n+1=\left(n^3+\frac{n}{2}-1\right)^2$$

And thus since your expression lies between two consecutive squares, it cannot be a square.

Answer 2

Hint:

for $n>2,$

$ n^3+\dfrac{n-1}2<\sqrt{n^6+n^4+1}<n^3+\dfrac n2.$

Answer 3

The only value that works is $n=2$.

Indeed, if there is another value that works, then there exists integers $n > 2$ and $k>0$ such that

$$n^6 + n^4 +1 = (n^3+k)^2$$

which implies that there exists exists integers $n > 2$ and $k>0$ such that

$$n^4+1 = 2n^3+k^2$$

However, note that on the one hand, for $k=n/2$ that $2n^3+k^2 = n^4+n^2/4$ which is strictly greater than $n^4+1$ for $n > 2$ [and equal to $n^4+1$ for $n=2$]. On the other hand, or $k=n/2-1$, note that $2n^3+k^2 < n^4-n^3+n^2/4 < n^4+1$ for $n > 2$.

Can you see how this implies no integral $k$ will solve the equation $n^4+1 = 2n^3+k^2$, and then in turn $n^4+1 = 2n^3+k^2$ for integral $n >2$?