Question

Prove that the product of four consecutive natural numbers cannot be a perfect cube.

Answer 1

Consider the product
$P=n(n+1)(n+2)(n+3)$, where n is a natural number.
If possible, that P is perfect cube $=k^3$
Two cases arises.
Case-I: If 'n' is odd.
$n,(n+1),(n+3)$ are all prime to $n+2$
Now, we know that every common divisor of $n+p$ and $n+q$ must divide $q-p$
$\therefore n+2$ and $n(n+1)(n+3)$ are relatively prime.
Since, their product is a perfect cube, each of them must be a perfect cube.
Since $n^3<n(n+1)(n+3)<(n+3{)}^3$
$\therefore n(n+1)(n+3)=(n+1{)}^{3}or(n+2{)}^3$
As $n(n+1)(n+3)and(n+2{)}^2$ are relatively prime, so second posibility ruled out.
Also $n(n+1)(n+3)=(n+1{)}^3\Rightarrow n=1$.
Since $P=24,$ when $n=1$ which is not a perfect cube. So the possibility $n=1$ is also ruled out. So n cannot odd.
Case-II: If 'n' is even.
Then $n+1$ is prime to $n,n+2andn+3$. Consequency $n+1$ is relatively prime to $n(n+2)(n+3).$ Since the product of relatively prime number $n+1andn(n+2)(n+3)$ is a perfect cube, each of them must be perfect cube.
$n^3<n(n+2)(n+3)<(n+3{)}^3$
$\therefore n(n+2)(n+3)=either(n+1{)}^3$
or $(n+2{)}^{3}sincen(n+2\left)\right(n+3)andn+1$ are relatively prime.
$\therefore$ First possibility ruled out.
Also $n(n+2)(n+3)=(n+2{)}^3$
$\Rightarrow n+4=0$
which is out of question. Consequency n cannot be even.