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Hazel of Madras College, St Andrew's, Fife and
David of Reading School, Berkshire both sent in good proofs of the
first result. Hazel worked out some examples of her own and then
made the conjecture that
$2(x^2 + y^2) = (x + y)^2 + (x - y)^2$
Hazel's proof was as follows:
$(x + y)^2 + (x - y)^2 = (x + y)(x + y) + (x - y)(x - y)$
$= x^2 + 2xy + y^2 + x^2 - 2xy + y^2$
$= 2x^2 + 2y^2$
$= 2(x^2 + y^2)$
So double the sum of two squares is always equal to the sum of
two squares.
Robert of Newcastle-Under-Lyme School provided
an excellent solution and managed to crack the second toughnut part
of the problem .
Robert sensibly looked at some special cases
with low numbers before making a conjecture that
$3(x^2+y^2+z^2) = (x+y+z)^2 +(z-x)^2 +(z-y)^2 +(y-x)^2$
His proof of the result was:
L.H side = $3x^2 + 3y^2 + 3z^2$
R.H side = $(x + y + z)^2 + (z^2 -2xz + x^2) + (z^2 - 2yz +
y^2)+ (y^2 -2xy + x^2)$
$= (x + y + z)^2+2z^2 + 2y^2 + 2x^2 - 2xz - 2yz -2xy $
$ =(x^2 + y^2 +z^2 +2xy +2yz + 2xz)+ 2z^2 + 2y^2 + 2x^2 -2xz-2yz
- 2xy $
$ = 3x^2 + 3y^2 + 3z^2 $
This proves the result.
Finally, Simon of Elizabeth College,
Guernsey managed to prove an extended general version of the result
incorporating the sum of n squares. This uses a lot of summations
and notations that you would usually only encounter at A-level, so
this part is for 16 and above!
Consider the case with n squares, say $a_1^2, a_2^2, \ldots, a_n^2$
multiplied by n equalling $x$ other squares.
Consider $$\left(\sum_{i=1}^{n} a_i\right)^2$$
$$\left(\sum_{i=1}^{n} a_i\right)^2 = 2\sum_{i=1}^{n}
\sum_{j=i+1}^{n} a_ia_j + \sum_{i=1}^{n} a_i^2$$
Now consider $$ \sum_{r=1}^{n} {\sum_{s=r+1}^{n}
(a_r-a_s)^2}$$
$$= \sum_{r=1}^{n} {\sum_{s=r+1}^{n} ( a_r^2 - 2a_ra_s + a_s^2
)}$$
$$= \sum_{r=1}^{n} {\sum_{s=r+1}^{n} a_r^2} - 2\sum_{r=1}^{n}
{\sum_{s=r+1}^{n} a_ra_s} + \sum_{r=1}^{n} {\sum_{s=r+1}^{n}
a_s^2}$$
Adding our first two functions $$\left(\sum_{i=1}^{n} a_i\right)^2
+ \sum_{r=1}^{n} {\sum_{s=r+1}^{n} (a_r-a_s)^2}$$
$$=\sum_{i=1}^{n}a_i^2+\sum_{r=1}^{n}{\sum_{s=r+1}^{n}
a_r^2}+\sum_{r=1}^{n}{\sum_{s=r+1}^{n}a_s^2}$$
Having shown that the `double' summation removes the extra terms in
the form of $2a_ra_s$, I now need to show that $$n \left (
\sum_{r=1}^n {a_r^2} \right ) =
\sum_{i=1}^{n}a_i^2+\sum_{r=1}^{n}{\sum_{s=r+1}^{n}
a_r^2}+\sum_{r=1}^{n}{\sum_{s=r+1}^{n}a_s^2}$$
Focusing at the moment on
$$ \sum_{r=1}^{n}{\sum_{s=r+1}^{n}
a_r^2}+\sum_{r=1}^{n}{\sum_{s=r+1}^{n}a_s^2} $$
we can see that for $n=1$ there will be $n-1$ terms coming from the
left summation and none from the right. Looking at $n=2$ there will
be $n-2$ terms on the left, and there will be one term on the
right, which is $n-1$ in total. Now looking at the $n^{th}$ term we
can see that there will be $n-n$ terms on the left and $n-1$ on the
left which is $(n-n)+(n-1)=n-1$ in total. Therefore, we can say
that
$$ \sum_{i=1}^{n}a_i^2+\sum_{r=1}^{n}{\sum_{s=r+1}^{n}
a_r^2}+\sum_{r=1}^{n}{\sum_{s=r+1}^{n}a_s^2} $$
$$ = \sum_{i=1}^{n}a_i^2 + (n-1)\sum_{r=1}^{n}a_r^2 $$
$$ = n \left ( \sum_{r=1}^n {a_r^2} \right ) $$
Which is what we were trying to prove earlier.
To finalise the proof, we need to know how many terms there will be
which equal the n multiplied by n square numbers. To to this I have
summed the number of bracketed pairs.
$$\sum_{r=1}^{n} {\sum_{s=r+1}^{n} 1}$$
$$= \sum_{r=1}^{n} ( n - ( r + 1 ) + 1 )$$
$$= \sum_{r=1}^{n} ( n -r )$$
$$= \sum_{r=1}^{n} n - \sum_{r=1}^{n} r$$
$$= n^2 - \frac{n(n+1)}{2}$$
Since $T_{n-1} + T_{n} = n^2$, where $T_n$ is the $n^{th}$ triangle
number, and $\frac{n(n+1)}{2}$ is the $n^{th}$ triangle number we,
can see that
$$\sum_{r=1}^{n} {\sum_{s=r+1}^{n} 1} = T_{n-1}$$
Therefore, there will be $T_{n-1}$ bracketed pairs, plus the one
large square, which is $T_{n-1}+1$ in total. Therefore, one can say
that.
For n square numbers, multiplied
by $n$, there will be $T_{n-1}+1$ other square numbers which equal
the same value.
We can use this to determine how many `terms' there will be in our
final equation. We can see that there will be $T_{n-1}+1$ terms,
remembering to add the `largest' term containing all of the other
terms. This leads us to the conclusion that $$n \left (
\sum_{r=1}^n {a_r^2} \right ) = \left(\sum_{i=1}^{n} a_i\right)^2 +
\sum_{r=1}^{n} {\sum_{s=r+1}^{n} (a_r-a_s)^2}$$ which can be
expressed in words as n square numbers, multiplied by $n$, will be
equal to $T_{n-1}+1$ other square numbers.