3. It is possible to write $15129=123^{2}$ as the sum of three distinct squares: $15129=$ $27^{2}+72^{2}+96^{2}$
(i) By using the identity $(a+b)^{2} \equiv a^{2}+2 a b+b^{2}$, or otherwise, find another way to write 15129 as the sum of three distinct squares.
(ii) Hence, or otherwise, show that $378225=615^{2}$ can be written as the sum of six distinct squares.
(i) Writing 123 as $121+2$, we can see that
$$
15129=123^{2}=(121+2)^{2}=121^{2}+2(121)(2)+2^{2} .
$$

Now,
$$
2 \times 121 \times 2=2 \times 2 \times 121=2^{2} \times 11^{2}=22^{2} \text {. }
$$

So
$$
15129=121^{2}+22^{2}+2^{2} .
$$

Comment
The above solution would earn all the marks, but doesn't really help see how one might discover it. So what might the thought process that leads to this solution look like?

Prompted by the given identity, we might realise it would suffice to find two positive integers $a$ and $b$ such that $a+b=123$ and $2 a b$ is a square.
Trying a few numbers, you might stumble upon 121 and 2 by chance (one of them is very small which makes that more likely), or perhaps you might realise that if you can find $a$ and $b$ with $a$ a square and $b$ twice a square then that would work, since $2 a b$ would be $2^{2} \times$ square $\times$ square, which is itself square. Choosing $b=2 \times 1$ (the smallest square) finds you the answer.
There is, in fact, a second solutions you could find using this approach: $98^{2}+70^{2}+25^{2}$ (when $a=25$ and $b=2 \times 49$ ). .

REMARK
There are in fact 23 ways (ignoring reordering) to write 123 as the sum of the squares of three distinct positive integers.
\begin{tabular}{ccccc}
$(2,22,121)$ & $(2,55,110)$ & $(7,14,122)$ & $(7,34,118)$ & $(7,62,106)$ \\
$(7,74,98)$ & $(12,36,117)$ & $(12,72,99)$ & $(14,73,98)$ & $(22,71,98)$ \\
$(22,82,89)$ & $(23,26,118)$ & $(23,50,110)$ & $(23,58,106)$ & $(25,70,98)$ \\
$(26,62,103)$ & $(27,72,96)$ & $(34,58,103)$ & $(36,72,93)$ & $(41,62,98)$ \\
$(50,55,98)$ & $(58,62,89)$ & $(58,71,82)$ & &
\end{tabular}
(ii) Note that $615=5 \times 123$. Hence $615^{2}=5^{2} \times 123^{2}$. Since $5^{2}=3^{2}+4^{2}$, we have
$$
\begin{aligned}
615^{2} & =\left(3^{2}+4^{2}\right) \times 123^{2} \\
& =3^{2} \times 123^{2}+4^{2} \times 123^{2} \\
& =3^{2} \times\left[2^{2}+22^{2}+121^{2}\right]+4^{2} \times\left[2^{2}+22^{2}+121^{2}\right] \\
& =(3 \times 2)^{2}+(3 \times 22)^{2}+(3 \times 121)^{2}+(4 \times 2)^{2}+(4 \times 22)^{2}+(4 \times 121)^{2} \\
& =6^{2}+66^{2}+363^{2}+8^{2}+88^{2}+484^{2}
\end{aligned}
$$