C4 For any positive integer $n$, an $n$-tuple of positive integers $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is said to be supersquared if it satisfies both of the following properties:
(1) $x_{1}>x_{2}>x_{3}>\cdots>x_{n}$.
(2) The sum $x_{1}^{2}+x_{2}^{2}+\cdots+x_{k}^{2}$ is a perfect square for each $1 \leq k \leq n$.
For example, $(12,9,8)$ is super-squared, since $12>9>8$, and each of $12^{2}, 12^{2}+9^{2}$, and $12^{2}+9^{2}+8^{2}$ are perfect squares.
(a) (2 marks) Determine all values of $t$ such that $(32, t, 9)$ is super-squared.
(b) (2 marks) Determine a super-squared 4-tuple $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ with $x_{1}<200$.
(c) (6 marks) Determine whether there exists a super-squared 2012-tuple.
Solution: The only answer is $t=24$.
Note that $32^{2}+t^{2}=1024+t^{2}$ and $32^{2}+t^{2}+9^{2}=1105+t^{2}$ are perfect squares. Then there exist positive integers $a, b$ such that
$$
\begin{array}{l}
1024+t^{2}=a^{2} \\
1105+t^{2}=b^{2} .
\end{array}
$$

Subtracting the first equation from the second equation gives
$$
b^{2}-a^{2}=81 \Rightarrow(b-a)(b+a)=81 .
$$

The only ways 81 can be written as the product of two distinct positive integers is $81=1 \times 81$ and $81=3 \times 27$.

If $(b-a, b+a)=(1,81)$, then $b-a=1$ and $b+a=81$. Summing these two equations yield $2 b=82$. Therefore, $b=41$. Hence, $a=40$. Therefore, $t^{2}=a^{2}-32^{2}=40^{2}-32^{2}=$ $8^{2}\left(5^{2}-4^{2}\right)=8^{2} \cdot 3^{2}$. Hence, $t=24$.

We now verify that $(32,24,9)$ is indeed super-squared. Clearly, the tuple is strictly decreasing, i.e. satisfies condition (1). Finally, $32^{2}+24^{2}=8^{2}\left(4^{2}+3^{2}\right)=8^{2} \cdot 5^{2}=40^{2}$ and $32^{2}+24^{2}+9^{2}=40^{2}+9^{2}=1681=41^{2}$. Therefore, the tuple also satisfies condition (2).

If $(b-a, b+a)=(3,27)$, then $b-a=3$ and $b+a=27$. Summing these two equations gives $2 b=30$. Therefore, $b=15$. Hence, $a=12$. Therefore, $t^{2}=a^{2}-32^{2}=12^{2}-32^{2}x_{2}>\cdots>x_{n}$, $a x_{1}>a x_{2}>\cdots a x_{n}$. Since $x_{1}^{2}+x_{2}^{2}+\cdots+x_{k}^{2}$ is a perfect square, $x_{1}^{2}+x_{2}^{2}+\cdots+x_{k}^{2}=m^{2}$ for some positive integer $m$. Therefore, $\left(a x_{1}\right)^{2}+\cdots+\left(a x_{k}\right)^{2}=(a m)^{2}$. Hence, $\left(a x_{1}, \cdots, a x_{n}\right)$ is super-squared.

From the example in the problem statement, $(12,9,8)$ is super-squared. Therefore, $12(12,9,8)=(144,108,96)$ is also super-squared. Note that $12^{2}+9^{2}+8^{2}=17^{2}$. Hence, $144^{2}+108^{2}+96^{2}=204^{2}=12^{2} \cdot 17^{2}$.

Note that $13^{2} \cdot 17^{2}=\left(12^{2}+5^{2}\right) \cdot 17^{2}=12^{2} \cdot 17^{2}+5^{2} \cdot 17^{2}=12^{2} \cdot 17^{2}+85^{2}$. Therefore, $221^{2}=13^{2} \times 17^{2}=144^{2}+108^{2}+96^{2}+85^{2}$. And so we conclude that $(144,108,96,85)$ is super-squared.

Comment: The list of all super-squared 4 -tuples $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ with $x_{1}b r$, or equivalently, $a / b>r / x_{k}$. Note that $r, x_{k}$ are determined from the tuple $\left(x_{1}, \cdots, x_{k}\right)$. Hence, it suffices to show that there exists a Pythagorean triple $(a, b, c)$, with $a^{2}+b^{2}=c^{2}$ such that $a / b>r / x_{k}$. In general, we need to show that $a / b$ can be arbitrarily large.

Note that $(a, b, c)=\left(m^{2}-1,2 m, m^{2}+1\right)$ is a Pythagorean triple for any positive integer $m$. This is clear since $\left(m^{2}-1\right)^{2}+(2 m)^{2}=m^{4}-2 m^{2}+1+4 m^{2}=m^{4}+2 m^{2}+1=\left(m^{2}+1\right)^{2}$. In such a case,
$$
\frac{a}{b}=\frac{m^{2}-1}{2 m}=\frac{m}{2}-\frac{1}{2 m}>\frac{m}{2}-1,
$$
which can be made arbitrarily large. This completes the induction proof.