41st Balkan Mathematical Olympiad

Suppose otherwise and let $m\in \mathbb{Z}$ be such that $n^4-12n^2+144=m^3$. Firstly, $m$ is clearly positive and we can assume that $n$ is a positive integer (since $n=0$ clearly doesn't work). Note that the polynomial $x^4-12x^2+144$ may be factored as $$x^4-12x^2+144=(x^2+12{)}^2-(6x{)}^2=(x^2-6x+12)(x^2+6x+12).$$ By repeatedly applying Euclid's algorithm, we get $$\gcd (n^2-6n+12,n^2+6n+12)=\gcd (n^2-6n+12,12n),(1)$$ and then $$\gcd (n^2-6n+12,n)=\gcd (n^2-6n+12-(n-6)n,n)=\gcd (12,n).(2)$$ We now distinguish three cases.

Case I. $n$ is even. Write $n=2k$ for some $k\in \mathbb{N}$ and we have $16(k^4-3k^2+9)=m^3$. Hence $m$ is divisible by 4 which means that $m^3$ is divisible by 64. But $k^4-3k^2+9$ is always odd, so $16(k^4-3k^2+9)$ cannot be divisible by 64, a contradiction.

Case II. $n$ is divisible by 3. Write $n=3l$ for some $l\in \mathbb{N}$ and we have $9(9l^4-3l^2+16)=m^3$. Hence $m$ is divisible by 3 which means $m^3$ is divisible by 27. But $9l^4-3l^2+16$ is not divisible by 3, so $9(9l^4-3l^2+16)$ cannot be divisible by 27, a contradiction.

Case III. Suppose that $\gcd (n,6)=1$. Then $\gcd (n^2-6n+12,n^2+6n+12)=1$ by (1) and (2) (since $\gcd (n^2-6n+12,12)=\gcd (n(n-6),12)=1$, because $\gcd (n,6)=1$), thus $n^2-6n+12=(n-3{)}^2+3$ and $n^2+6n+12=(n+3{)}^2+3$ are both perfect cubes of integers. However, we get a contradiction with the following lemma.

Lemma 1. For every even integer $x$, the number $x^2+3$ is not a perfect cube of an integer.

Proof. Suppose otherwise, namely that there exists a positive integer $y$ such that $x^2+3=y^3$. The last relation modulo 4 gives $y\equiv -1(\mathrm{mod}4)$. The equation then turns to $$x^2+2^2=y^3+1=(y+1)(y^2-y+1).$$ But we have $y^2-y+1\equiv (-1{)}^2-(-1)+1\equiv -1(\mathrm{mod}4)$, hence there exists a prime number $p\equiv -1(\mathrm{mod}4)$ such that $p\mid y^2-y+1\mid x^2+2^2$. But it is well known that from here we should have $p\mid x$ and $p\mid 2$. This means that $p=2$, which contradicts $p\equiv -1(\mathrm{mod}4)$. □

We conclude that no such integer $m$ can exist, thus $n^4-12n^2+144$ is never a perfect cube of an integer. □