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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:
$$ x^2 +3x + 1$$
$$ \mbox{So} x^4 +6x^3 +11x^2 +6x +1 = (x^2 +3x +1)^2 $$
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.