prove that [p!/(p-4)!] + 1 is a perfect square for all natural p.

Question

one can observe that $[p!/(p-4)!] + 1$ is basically the product of four consecutive integers plus one.Since this is $$ \begin{eqnarray} p(p+1)(p+2)(p+3)+1 & = &(p^2+3p)(p^2+3p+2)+1 \\ & = & [(p^2+3p+1)−1][(p^2+3p+1)+1]+1 \\ & = & (p^2+3p+1)^2 \end{eqnarray} $$ which is a perfect square for all natural $p$ but my teacher says that there is a more elegant method of doing it, help in any form would be appreciated.

Answer 1

Only thing I can see:

Note that $p(p-3)=p^2-3p$ and $(p-1)(p-2)=p^2-3p+2$. It follows that we can write your expression as $$(p^2-3p)^2+2(p^2-3p)+1=m^2+2m+1=(m+1)^2$$ Where we have introduced $m=p^2-3p$ ,

Answer 2

Slightly different approach... where $p=q+3$

$$S_q=\frac{(p)!}{(p-4)!}+1=\frac{(q+3)!}{(q-1)!} + 1=q(q+1)(q+2)(q+3)+1$$

Let $q=a-\frac{3}{2}$

Then

$$S_q=\left(a-\frac{3}{2}\right)\left(a-\frac{1}{2}\right)\left(a+\frac{1}{2}\right)\left(a+\frac{3}{2}\right)+1$$

$$S_q=\left(a^2-\frac{3^2}{2^2}\right)\left(a^2-\frac{1}{2^2}\right)+1$$

$$S_q=a^4-\frac{10}{4}a^2+\frac{3^2}{4^2}+1$$

$$S_q=a^4-\frac{10}{4}a^2+\frac{5^2}{4^2}$$

$$S_q=\left(a^2-\frac{5}{4} \right)^2$$

$$S_q=\left((q+3/2)^2-\frac{5}{4} \right)^2$$

$$S_p=\left((p-3/2)^2-\frac{5}{4} \right)^2$$