Show that $\sqrt{1+n(n+1)(n+2)(n+3)}$ is a whole number for all whole numbers $n$.

Show that $\sqrt{1+n(n+1)(n+2)(n+3)}$ is a whole number for all whole numbers $n$.

I can see that there are four consecutive numbers, meaning that the expression can be written as $\sqrt{1+24m}$ Also, it is easy to see that the expression is true for $n=1$ but I can't get the induction to work...

Notice that

$$1+n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n+1=(n^2+3n)^2+2(n^2+3n)+1$$

Let $n^2+3n=t$

then we have $$1+n(n+1)(n+2)(n+3)=t^2+2t+1=(t+1)^2$$

so $$ \sqrt{1+n(n+1)(n+2)(n+3)} =t+1=n^2+3n+1$$ which is clearly a whole number for $n \in Z$

Another approach is to note that $(n+1)(n+2)=n(n+3)+2$, so if $M=n(n+3)+1$ then $M-1=n(n+3)$ and $M+1=(n+1)(n+2)$ so $M^2-1=(M-1)(M+1)=n(n+1)(n+2)(n+3)$.