Consider the equation $x^{2}-2 y^{2}=1$, which we label (1). There are many pairs of positive integers $(x, y)$ that satisfy equation (1).

(a) Determine a pair of positive integers $(x, y)$ with $x \leq 5$ that satisfies equation (1).

(b) Determine a pair of positive integers $(u, v)$ such that

$$
(3+2 \sqrt{2})^{2}=u+v \sqrt{2}
$$

and show that $(u, v)$ satisfies equation (1).

(c) Suppose that $(a, b)$ is a pair of positive integers that satisfies equation (1).

Suppose also that $(c, d)$ is a pair of positive integers such that

$$
(a+b \sqrt{2})(3+2 \sqrt{2})=c+d \sqrt{2}
$$

Show that $(c, d)$ satisfies equation (1).

(d) Determine a pair of positive integers $(x, y)$ with $y>100$ that satisfies equation (1).
(a) With some trial and error, we find that the pair $(x, y)=(3,2)$ satisfies the given equation, since $3^{2}-2\left(2^{2}\right)=1$.

If we were asked to justify that this was the only solution or if we wanted a more systematic way to find this pair, we could try the possible values of $x$ from 1 to 5 and try to find one that gives a positive integer value of $y$.

If $x=1$, then $1-2 y^{2}=1$ or $2 y^{2}=0$, which does not give any positive integer solutions.

If $x=2$, then $4-2 y^{2}=1$ or $2 y^{2}=3$, which does not give any positive integer solutions.

If $x=3$, then $9-2 y^{2}=1$ or $2 y^{2}=8$, which has $y=2$ as a solution.

If $x=4$, then $16-2 y^{2}=1$ or $2 y^{2}=15$, which does not give any positive integer solutions.

If $x=5$, then $25-2 y^{2}=1$ or $2 y^{2}=24$, which does not give any positive integer solutions.

Therefore, the only pair of positive integers that satisfies equation (1) and has $x \leq 5$ is $(x, y)=(3,2)$.

(b) Expanding $(3+2 \sqrt{2})^{2}$, we obtain

$$
u+v \sqrt{2}=(3+2 \sqrt{2})(3+2 \sqrt{2})=9+6 \sqrt{2}+6 \sqrt{2}+8=17+12 \sqrt{2}
$$

Therefore, $(u, v)=(17,12)$ satisfies the equation $(3+2 \sqrt{2})^{2}=u+v \sqrt{2}$.

Furthermore, if $(u, v)=(17,12)$, then $u^{2}-2 v^{2}=17^{2}-2\left(12^{2}\right)=289-2(144)=1$, so $(u, v)=(17,12)$ satisfies equation (1), as required.

(c) Since $(a, b)$ satisfies (1), then $a^{2}-2 b^{2}=1$.

Since $(a+b \sqrt{2})(3+2 \sqrt{2})=c+d \sqrt{2}$, then

$$
c+d \sqrt{2}=3 a+2 a \sqrt{2}+3 b \sqrt{2}+2 b(2)=(3 a+4 b)+(2 a+3 b) \sqrt{2}
$$

and so $(c, d)=(3 a+4 b, 2 a+3 b)$.

(It is not hard to see that if $(c, d)=(3 a+4 b, 2 a+3 b)$, then we have

$$
c+d \sqrt{2}=(3 a+4 b)+(2 a+3 b) \sqrt{2}
$$

To formally justify that $c+d \sqrt{2}=(3 a+4 b)+(2 a+3 b) \sqrt{2}$ implies $(c, d)=(3 a+4 b, 2 a+3 b)$, we can rearrange the equation to obtain

$$
c-3 a-4 b=(2 a+3 b-d) \sqrt{2}
$$

The left side of this equation is an integer. If $2 a+3 b-d \neq 0$, the right side of this equation is irrational (because $\sqrt{2}$ is irrational), so cannot equal an integer. Thus, $2 a+3 b-d=0$ or $d=2 a+3 b$. This implies that $c-3 a-4 b=0$ or $c=3 a+4 b$.)

To show that $(c, d)$ satisfies equation (1), we need to show that $c^{2}-2 d^{2}=1$ :

$$
\begin{aligned}
c^{2}-2 d^{2} & =(3 a+4 b)^{2}-2(2 a+3 b)^{2} \\
& =\left(9 a^{2}+24 a b+16 b^{2}\right)-2\left(4 a^{2}+12 a b+9 b^{2}\right) \\
& =9 a^{2}+24 a b+16 b^{2}-8 a^{2}-24 a b-18 b^{2} \\
& =a^{2}-2 b^{2} \\
& =1 \quad\left(\text { since } a^{2}-2 b^{2}=1\right)
\end{aligned}
$$

Therefore, $(c, d)$ satisfies equation (1), as required.
(d) From (c), we know that if $(a, b)$ is a solution to equation (1), then $(c, d)=(3 a+4 b, 2 a+3 b)$ is also a solution to equation (1).

From (b), we know that $(17,12)$ is a solution to equation (1).

We use these two facts together.

Since $(17,12)$ is a solution, then $(3(17)+4(12), 2(17)+3(12))=(99,70)$ is a solution.

Since $(99,70)$ is a solution, then $(3(99)+4(70), 2(99)+3(70))=(577,408)$ is a solution, which has $y>100$.

We can verify that $577^{2}-2\left(408^{2}\right)=1$, as required.