# Show that $3n^4+3n^2+1$ is never a perfect square [duplicate]

I am looking for a proof for the fact that $$3n^4+3n^2+1$$ can never be a perfect square for a natural number $$n>0$$.

I know for a fact that the statement must be true as it came up as one of the cases in a solution of the diophantine equation $$y^2=x^3-1$$ using the LTE lemma and, according to two different solutions I have come across, this equation has no solutions apart form $$(x,y)=(1, 0)$$.

I have spent a considerable amount of time looking for a suitable attack strategy, however, I was not able to make any progress. Can anyone help?

• What attack strategies did you try? – Arthur Oct 29 '20 at 20:02
• As $n=0$ works, no approach that uses only modular arithmetic can work. Hence, it must have something to do with bounding. To that end, I tried to factor $m^2=3n^4+3n^2+1$ and get some useful divisibilities (which always also lead to inequalities, of course) or to set $m=n^2+r$ with $0<r<n^2$ and then see what happens to $r$. These approaches did not work out for me, though (which does of course not mean that they are doomed altogether). – mxian Oct 29 '20 at 20:07
• Here is a similar question – J. W. Tanner Oct 29 '20 at 20:22
• @Sil: how did you change $k^2$ to $k^3$? – J. W. Tanner Oct 29 '20 at 20:38
• @J.W.Tanner Using approach0 – Sil Oct 29 '20 at 20:46

$$3n^4+3n^2+1-k^2 =0\implies \Delta = 9-12(1-k^2)$$

Since $$\Delta$$ is divisible by $$3$$ and it is a perfect square we have $$3d^2 = 4k^2-1$$

Now $$2k-1$$ and $$2k+1$$ are relatively prime so we have $$2k-1=3a^2\;\;\;\wedge \;\;\;\; 2k+1=b^2$$ or $$2k-1=a^2\;\;\;\wedge \;\;\;\; 2k+1=3b^2$$

In first case we get $$2 =b^2-3a^2$$ and in second $$2=3b^2-a^2$$ (which has no solution, see by modulo 3). So we are left with $$2 =b^2-3a^2$$ which has something to do with the Pell equation...

Suppose that $$3n^4+3n^2+1=m^2$$, so that $$3n^4+3n^2+(1-m^2)=0$$ This is a quadratic in $$n^2$$. $$n^2=\frac{-3\pm\sqrt{9-12(1-m^2)}}{6}=-\frac12\pm\frac{\sqrt{12m^2-3}}6$$

So if $$n$$ is an integer, there is an integer $$k$$ such that $$12m^2-3=k^2$$ Clearly, $$3|k$$, so let $$k=3t$$ and let $$s=2m$$. Then we have $$3s^2-9t^2=3\\s^2-3t^3=1$$ a Pell's equation.

This is as far as I've carried it, but presumably, solving for $$s$$ and $$t$$ and putting it back in terms of $$m$$ and $$k$$ will show that you can't have an integer solution to the original equation.

$$3n^4 + 3n^2 + 1 - x^2=0$$ is an elliptic curve with Weierstrass form $$s^3 + t^2 - 15s + 22=0$$. According to Sage, the rank is $$0$$ and the torsion group has order $$2$$, consisting of a point at infinity (which corresponds to $$n=0, x=-1$$) and $$s=2,t=0$$ (which corresponds to $$n=0,x=1$$). Thus $$n=0$$, $$x = \pm 1$$ are the only rational points.

• Fair enough, I am sure that this is a valid solution, but I am lacking the theoretical background to understand it and was also hoping that there would be a somewhat elementary solution. Thank you nevertheless! – mxian Oct 29 '20 at 20:22
• @mxian: See S.Dolan’s wonderful [elementary] solution here: math.stackexchange.com/a/4004298/93271 – Kieren MacMillan Jan 30 at 17:35