Square numbers in the form of $n^6 + n^4 +1$

Question

For what natural numbers is $n^6 + n^4 +1$ is a square of a natural number?

I just want to know if my way of solving it is correct or not. (I have checked the other threads regarding the same question but their answers have a different method than mine)

Here's what I did:

$$n^6 + n^4 +1 = n^6 + n^3\cdot n +1$$ $$=n^6 + \frac{2n^3\cdot n}{2} + 1$$ $$=n^6 + 2n^3\cdot \frac{n}{2} +1$$

This expression would be in the form of $(n^3 + 1)^2$ if $\frac{n}{2}=1$ and it would be in the form of $(n^3-1)^2$ if $\frac{n}{2}=-1$. Since $n\in \mathbb{N}$, $n=+2$

Is the way I did correct? Help would be appreciated.

Answer 1

You can't really say it would be the in form of $(n^3+1)^2$. Instead you have to show this.

So let this be in form of $(an^3+b)^2$

$$ a^2n^6+b^2+2abn^3=n^6+ 1 + n^4$$

If we compare coefficients,

$$ a^2n^6=n^6 => a=1 $$

$$ b^2=1 => b=1 $$

See we have neglected the $-1$ value for b because $-1$ isn't natural.

Now if we compare the last coefficent,

$$ 2abn^3=n^4 $$

$$ 2n^3=n^4 $$

$$ n=2$$