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This was a very popular problem with lots of solutions sent in. An impressive performance from many of you. The choice of names to list and solutions to show was a difficult one but I have chosen ones which give a spread of approaches and demonstrate that there is more than one way to crack a nut!

Tiffany of Island School offered this solution, similar to those sent in by many of you:

If you let the first of any 4 consecutive numbers be $x$, their product $+1$ will be $$ x(x+1)(x+2)(x+3) + 1 = x^4 +6x^3 +11x^2 +6x +1. $$ By trial and error, and inspection, I found the square root of the above expression: $$ x^2 +3x + 1$$ So

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

This should prove that the product of any $4$ consecutive numbers add $1$ is always a square number!

Daniel from Wales High School offered the following. I liked this because he made an effort to explain every step:

Since consecutive numbers follow on from each other, they are 1 apart from each other e.g. $1, 2, 3, 4, 5$.

So algebraic consecutive numbers follow the pattern $x, x+1, x+2, x+3, x+4 \dots$

We only want the first $4$: $x, x+1, x+2, x+3$. Now we need to times them all together and add $1$:

$$\begin{eqnarray} x(x + 1) &=& x^2 + x\\ (x^2 + x)(x + 2) &=& x^3 +3x^2 + 2x\\ (x^3 + 3x^2 + 2x)(x + 3) &=& (x^4 + 6x^3 + 11x^2 + 6x)\\ \end{eqnarray}$$

Adding $1$ gives us $x^4 + 6x^3 + 11x^2 + 6x + 1$

Now we have to prove that the above bracket is a perfect square.

If this factorises into $2$ equal brackets, then the rule is true.

Firstly we need to square root $x^4$. $x^2$ squared equals $x^4$ so our brackets look like this:

$$ (x^2 ...)(x^2 ... ) $$

The end part of the brackets must be the square root of $1$.

This could either be $1$ or $-1$ but because the terms we are factorising all include a $+$ sign we don't need to use $-1$.
The brackets are now as follows:
 
$$ (x^2+ ... +1)(x^2+ ... +1) $$
 
Finally the middle terms need to add to make $6x$ and times to make $9x^2$ (because $9x^2 + \mbox{ the } 2x^2$ already in the brackets).
 
So square root $9x^2$ and you get $3x$.
Also $3x+3x=6x$ so this is the number we are looking for: $(x^2 + 3x + 1)(x^2 + 3x + 1)=(x^2 + 3x + 1)^2$ which is a perfect square. QED

Haukur of King's School Canterbury saw a pattern in the mathematics and utilised it. Not the most concise solution but a lovely example of using pattern as a key to problem solving.

I noticed that in the examples given, the product of the first and last numbers was always one less than the squared number.
 
Similarly, the product of the second and third numbers was always one greater than the squared number.

To illustrate:

$\begin{eqnarray} 2\times3\times4\times5 +1 &=& 11^2 \\ 2\times5=10 &=& 11-1 \\ 3\times4=12 &=& 11+1 \\ \end{eqnarray}$

I used this as the basis for the proof.
I postulated that:
$n(n+1)(n+2)(n+3) +1 = [n(n+3) +1]^2 = [(n+1)(n+2) -1]^2 $
taking 1 from all three expressions leaves

$$ n(n+1)(n+2)(n+3) $$

$$(n(n+3) +1)^2 -1 $$

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

and if all three are equal, then the given statement that 'the product of four consecutive integers plus one is a perfect square' must be true, as
$[n(n+3) +1] \textbf{ and } [(n+1)(n+2) -1] $ (which can easily be shown to be equal) must always be an integer if n is an integer.
By expanding each into single polynomials, we find that they are all equal to
$n^4 + 6n^3 +11n^2 +6n $ Q.E.D.

A more elegant solution would have been to expand $n(n+1)(n+2)(n+3) +1 $, then factorise it into perfect squares.