Proving the product of four consecutive integers, plus one, is a square [duplicate]

Solution 1:

By the way, you mean $m+1 = x^4 + 6x^3 + 11x^2 + 6x + 1$.

Let's break it down. Obviously, you need a quadratic that, when squared, gives the above. How do you construct this? You need 3 numbers $a, b, c$ for $ax^2 + bx + c$.

Looking at the 4th degree thing above, what can you tell me about $a$, right off the bat? What about $c$? Then can you use this to tell you what $b$ is?

(I feel like I should not go further because I want you to solve it, but let me know if you need a clarification)

Solution 2:

To get a feel for the problem, let's work backwards, starting from a square number. Take some integer $n$.

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

OK, so it looks like $m$ has two factors whose difference is 2. Could it be that in our product of consecutive integers, the product of two of them is $n-1$ and of the other two is $n+1$?

Let's throw in a simple example: $1\times2\times3\times4$. Notice how $1\times4=4$ and $2\times3=6$.

What about $2\times3\times4\times5$? This time $2\times5=10$ and $3\times4=12$.

It looks like the product of the "outer" pair is $n-1$ and the product of the "inner pair" is $n+1$.

Now we know how to attack this.

Let $k$ be some integer and $m=k(k+1)(k+2)(k+3)$.

$\begin{align} m &= k(k+1)(k+2)(k+3) \\ &= (k+1)(k+2)\times(k(k+3)) \ \text{ (collecting inner and outer terms)}\\ &= (k^2 + 3k + 2) \times (k^2 + 3k) \\ &= ((k^2 + 3k + 1) + 1) ((k^2 + 3k + 1) - 1) \\ &= (k^2 + 3k + 1) ^ 2 - 1 \qquad \text{ (since }(a+b)(a-b)=a^2-b^2\text{)}. \end{align}$

Since $k$ is an integer, $k^2 + 3k + 1$ is an integer so $m+1$ is a perfect square.

Solution 3:

Write the product of the four consecutive integers starting at some $n-1$, so that $$m=(n-1)n(n+1)(n+2)+1,$$ and expand: \begin{align} m&=(n^2-1)(n^2+2n)+1=n^2(n^2-1)+2n(n^2-1)+1 \\ &=(n^2-1)^2+2n(n^2-1)+\not 1+ n^2{-}\!\not1 \\ &= \bigl((n^2-1)+n\bigr)^2. \end{align}

Solution 4:

I wrote about this problem at some length on MathEducators StackExchange: MESE 10736.

See Part II there for several approaches taken by students (most of which are covered by other answers here; but the presentation is somewhat different).

One of the methods mentioned there is observing the symmetry above around $x= -3/2 = -1.5$, but then using this to inform a substitution: let $z = x + 1.5$ so that we have:

$$x(x+1)(x+2)(x+3) = (z-1.5)(z-0.5)(z+0.5)(z+1.5) = (z^2 - 1.5^2)(z^2 - 0.5^2)$$

Noting that $1.5^2 = 2.25$ and $0.5^2 = 0.25$, we could use one more substitution of $w = z^2 - 2.25$ to rewrite the final expression above as $w(w+2) = w^2 + 2w$, from which the addition of $1$ yields $(w+1)^2$ as desired. One can now rewrite in terms of just $x$ to finish matters off.

I think that the idea of the symmetry here is an important takeaway; incidentally, the problem is also broached in an exploratory manner at the beginning of Paul Zeitz's (2006) The Art and Craft of Problem Solving as Example 1.2.1.