$a^3+3a^2+a$ is never a perfect square.

Question

Prove that no number of the form $ a^3+3a^2+a $, for a positive integer $a$, is a perfect square.

This problem was published in the Italian national competition (Cesenatico 1991). I've been trying to solve it using modular arithmetic, without success. Thank you for your interest in this question.

Answer 1

Well $a^3+3a^2+a=a(a^2+3a+1)$ and you can see that $\gcd(a,a^2+3a+1)=1$. So if it's a square, then $a^2+3a+1$ has to be a square too.

But that's not possible, because it's strictly between $(a+1)^2=a^2+2a+1$ and $(a+2)^2=a^2+4a+4$.

Answer 2

There is already an elementary proof here, but I just wanted to say that this equation defines an elliptic curve $E:y^2=x^3+3x^2+x$. The Mordell-Weil theorem states that the set of rational points is a finitely generated abelian group, so $$E(\mathbb{Q})\cong T_E \times \mathbb{Z}^{R_E},$$ where $T_E$ is a finite subgroup, and $R_E\geq 0$ is called the rank of the elliptic curve. There are methods to calculate $T_E$ and $R_E$ (such as the Nagell-Lutz theorem, or the method of $2$-descent), and these methods are implemented in software such as Sage and Magma. In this case: $$T_E=\{\mathcal{O}, (-1,-1), (0,0), (-1,1)\} \cong \mathbb{Z}/4\mathbb{Z},$$ where $\mathcal{O}$ is the point at infinity with projective coordinates $[0,1,0]$, and $R_E=0$, so there are no points of infinite order. So the set of all rational points (not just integral points) on $E$ is: $$E(\mathbb{Q})=\{\mathcal{O}, (-1,-1), (0,0), (-1,1)\}.$$

I calculated the torsion and rank using the online Magma calculator:

E:=EllipticCurve([0,3,0,1,0]);

Rank(E);

TorsionSubgroup(E);

Since all the points have $0$ or $-1$ as their $x$-coordinate, the desired result follows.