Prove that the product of four consecutive natural numbers are not the square of an integer

Question

Prove that the product of four consecutive natural numbers are not the square of an integer

Would appreciate any thoughts and feedback on my suggested proof, which is as follows:

Let $f(n) = n(n+1)(n+2)(n+3) $. Multiplying out the expression and refactoring it in a slightly different way gives $$f(n) = n^4 + 6n^3+11n^2+6n \\= n^4 + 6n^3 + 9n^2 + 2n^2 + 6n = (n^2 + 3n)^2 + 2n(n+3). \tag{1}\label{1} $$

We want to show that the only possible way for $ f(n) $ to be the square of an integer is if $ f(n) = (n^2 + 3n +1 ) ^2. $ We show this by proving that $ (n^2+3n)^2 < f(n) < (n^2+3n+2)^2 $. The left-hand side follows immediately from $(1)$, since $ 2n(n+3) > 0 $ for all $ n \geq 1 $, and the right-hand side can be verified by multiplying out both sides: $$ \begin{align} (n^2+3n)^2 + 2n(n+3) &< (n^2+3n+2)^2 \\ \iff n^4 + 6n^3 + 11n^2 + 6n &< n^4 + 9n^2 + 4 + 6n^3+4n^2+12n \\ \iff 0 &< 2n^2 + 6n + 4 \end{align} $$ which is true for all $n \geq 1 $. Now we note that $n^2+3n = n(n+3)$ is even since one of the factors $n$ or $n+3$ is even for all $n$. It follows that $ n^2+3n+1$ must be odd, and so $ (n^2+3n+1)^2 $ must be odd. But $ f(n) $ must be even, since either $n$ and $(n+2)$ is even, or $(n+1)$ and $(n+3)$ is even and an even number multiplied by an odd number is an even number. So $f(n) \neq (n^2 + 3n +1)$ and therefore $f(n)$ cannot be the square of an integer for all $n \geq 1 $.

Answer 1

Hint: $n(n+1)(n+2)(n+3) +1$ is a square.

Answer 2

Your solution is perfect and detailed.

Another alternative shorter approach based on observation is if we can consider $n(n+2)=k$ and then redefine the function as: $$\begin{align*}f(k) &= k(k+2)\\ f(k) &= (k+1)^2-1\end{align*}$$

Now, we need to prove that $f(k)$ is not the square of an integer. Well, there exists no two squares of a nonzero integer that differ by 1. Hope that helps!!