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Look at the end for a neat short cut
The solution below is based upon the one
submitted by Anna of Parkside school. I liked the explanation of
how Anna arrrived at the factorisation.
The youngest person to send in a solution was
Sairah of Kings Park School, Lurgan and Barinder sent an excellent
solution with lots of clear explanation.
Just three of the large number of solutions to
this problem. Well done to you all!
If $x_n = (n+1)^2 +(n+2)^2 + [(n + 1)(n + 2)]^2 $ then
\begin{eqnarray}x_n &=& n^2 + 2n + 1 +n^2 + 4n + 4 + [n^2 +
3n + 2]^2 \\ &=& 2n^2 + 6n + 5 + n^4 + 3n^3 + 2n^2 + 3n^3 +
9n^2 + 6n +2n^2 + 6n + 4 \\ &=& n^4 + 6n^3 + 15n^2 + 18n +
9 \end{eqnarray} If this can be made into a perfect square the
perfect square must have $n^2$ as it's highest term to get $n^4$
when you multiply out. The lowest term must be $3$ to give
$9$.
The middle term must be some number of $n$ ($an$).
$an\times n^2 \times 2$ must give $6n^3$
As the only parts of the square which can multiply together to
give $n^3$ are $n$ and $n^2$. $an\times n^2 \times 2 = 6n^3$, so
$2an=6n$, so $a=3$.
So the perfect square must be: $(n^2+3n+3)^2$.
To check, this multiplies out to:
$n^4+3n^3+3n^2+3n^3+9n^2+9n+3n^2+9n+9 =n^4+6n^3+15n^2+18n+9$.
$(n^2+3n+3)^2$ is a perfect square, and it is the solution for
$x_n$.
Now you don't actually need to worry
about the $n$ value matching the n of the left hand side . . .
.
You just need to establish that $ (n)^2 +(n+1)^2 + [n(n + 1)]^2 $
is always a square of something
Can you follow that through, expanding and simplifying until you
get: $ n^4 + 2n^3 + 3n^2 + 2n + 1$?
I'll leave you to decide what 'square'
that is, but you might think of
something inspired just from looking at the algebra, or maybe
calculate the first line from the problem (it should come to $49$)
and take it from there .
Look out for this kind of 'simpler
version' of the algebra when you are problem solving, it isn't
always easy to spot but it can save you some effort when you do
notice the possibility.