find the number of perfect squares of the form - $$n^6 + n^4+1$$ where n is a natural number.

I've reached this equation.

$$n^4(n^2+1)=(k+1)(k-1)$$ I thought this might have something to do with prime factorization but I don't seem to be getting anywhere. Am I on the right track or should I approach this question differently?


We do not have a solution for $n=1$, so $n\geq 2$. Because $(k-1, k+1)\leq 2$ we should have $k+1\geq \frac{n^4}{2}$ and $k-1\leq 2(n^2+1)$. $n^4-2(2(n^2+1)+2)=n^4-4n^2-8=(n^2-2)^2-12>0$ for $n\geq3$. That means we only need to check $n=2: 64+16+1=81=9^2$ - unique solution

Also, there is an alternative way. Considering mod 4 you can show that $n$ should be even. After that consider $(n^3+\frac{n}{2})^2=n^6+n^4+\frac{n^2}{4}> n^6+n^4+1$ for $n>2$ and $(n^3+\frac{n}{2}-1)^2=n^6+n^4-2n^3+\frac{n^2}{4}-n+1=n^6+n^4+1-(n^3-\frac{n^2}{4}-n)<n^6+n^4+1$ for $n>2$

So again you only need to check 2 cases.

  • $\begingroup$ In the first method, how did you get the two inequalities $k+1\geq \frac{n^4}{2}$ and $k-1\leq 2(n^2+1)$ $\endgroup$ – Karan May 15 at 9:48
  • 1
    $\begingroup$ Remember that $(k+1, k-1)=2$ or 1. That means that you cannot ''split" divisors(maybe except 2) of $n^4$ between factors $\endgroup$ – AO1992 May 15 at 9:55
  • $\begingroup$ Why though, lets say $p$ divides $n^4$ and I divide $n^4$ by $p^4$ and multiply that with $(n^2+1)$, that wont affect the GCD would it. $\endgroup$ – Karan May 16 at 16:06
  • $\begingroup$ Actually yeah, you are correct, my bad. $\endgroup$ – AO1992 May 16 at 16:27

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