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

Solution 1:

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.

Solution 2:

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

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

With this in mind

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

Solution 3:

Here's another way which begins by exploiting a symmetry in the expression.

Notice that if you substitute $x=p+\frac{3}{2}$, the expression becomes

$$\left(x-\frac{3}{2}\right)\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(x+\frac{3}{2}\right) + 1$$

Now see that the terms make the product of 2 differences of squares

$$\begin{align} & \quad \left(x+\frac{3}{2}\right)\left(x-\frac{3}{2}\right)\left(x+\frac{1}{2}\right)\left(x-\frac{1}{2}\right) + 1 \\&= \left(x^2-\frac{9}{4}\right)\left(x^2-\frac{1}{4}\right) + 1 \\ &= \left(x^4 - \frac{10}{4} x^2 + \frac{9}{16}\right) + 1 \\ &= x^4 - \frac{10}{4} x^2 + \frac{25}{16} \\ &= \left(x^2 - \frac{5}{4}\right)^2 \\ &= \left(p^2 + 3p + 1\right)^2 \end{align}$$

which is a perfect square.