Prove that the product of four consecutive positive integers plus one is a perfect square

Your technique should have worked, but if you don't know which expansions to do first you can get yourself in a tangle of algebra and make silly mistakes that bring the whole thing crashing down.

The way I reasoned was, well, I have four numbers multiplied together, and I want it to be two numbers of the same size multiplied together. So I'll try multiplying the big one with the small one, and the two middle ones.

$$p(p+1)(p+2)(p+3) + 1 = (p^2 + 3p)(p^2 + 3p + 2) + 1$$

Now those terms are nearly the same. How can we force them together? I'm going to use the basic but sometimes-overlooked fact that $xy = (x+1)y - y$, and likewise $x(y + 1) = xy + x$.

$$\begin{align*} (p^2 + 3p)(p^2 + 3p + 2) + 1 &= (p^2 + 3p + 1)(p^2 + 3p + 2) - (p^2 + 3p + 2) + 1 \\ &= (p^2 + 3p + 1)(p^2 + 3p + 1) + (p^2 + 3p + 1) - (p^2 + 3p + 2) + 1 \\ &= (p^2 + 3p + 1)^2 \end{align*}$$ Tada.

$(n-1)(n+1)+1 = n^{2}$.

Note that $(n+1)-(n-1)=2$.

With this in mind

$$\begin{align*} p(p+1)(p+2)(p+3)+1 &= (p^{2}+3p)(p^{2}+3p+2)+1 \\ &= [(p^{2}+3p+1)-1][(p^{2}+3p+1)+1]+1 \\ &= (p^{2}+3p+1)^2 \end{align*}$$

Below I present a generalization. $ $ Using the abbreviations$\rm\ \ c = a\!+\!b,\ \ \color{red}d = ab/2\:\ $ we compute

$$\rm\begin{eqnarray} &&\rm\qquad\quad\ \color{blue}{(x\!+\!a)\,(x\!+\!b)}\,(x\!+\!c)\,x &=&\rm\, \color{blue}{(x^2\!+cx\ \ +\ \ ab\ \ \ \, )}\,(x^2+cx\:\!) \\ && &=&\rm\, (x^2\!+cx+d\ \,\color{red}{+\, d})\,(x^2+cx+d\, \color{red}{-\,d}) \\ && &=&\rm\, (x^2\!+cx+d)^2\! \color{red}{- d^2} \\ \rm b=2\quad &\Rightarrow&\rm\quad\ \ \ \ x(x+a)(x\!+\!2)(x\!+\!a\!+\!2) &=&\,\rm (x^2\!+(a\!+\!2)\,x+a)^2 -a^2 \\ \rm a=1\quad &\Rightarrow&\rm\qquad\quad\ \ \ x(x\!+\!1)(x\!+\!2)(x\!+\!3) &=&\rm\, (x^2\!+3\:\!x+1)^2 -1\ \ \ as\ sought. \end{eqnarray}$$