Find all pairs of integers $x, y$ such that $2 x^{2}-6 x y+3 y^{2}=-1$.
We first characterize all solutions to the equation $a^{2}-3 b^{2}=1$ in nonnegative integers. This is a Pell equation, and the method used is standard. Suppose we have $(a+b \sqrt{3})(a-b \sqrt{3})=$ $a^{2}-3 b^{2}=1$ with $a>1$; then we can set $a^{\prime}=2 a-3 b, b^{\prime}=2 b-a$. Then $(a+b \sqrt{3})=(2+\sqrt{3})\left(a^{\prime}+b^{\prime} \sqrt{3}\right)$ and $a^{\prime 2}-3 b^{2}=\left(4 a^{2}-12 a b+9 b^{2}\right)-3\left(4 b^{2}-4 a b+a^{2}\right)=a^{2}-3 b^{2}=1$. Moreover, we claim $a^{\prime}$ and $b^{\prime}$ are also nonnegative, with $a^{\prime}0 ; 3 b^{2}=a^{2}-1>0 \Rightarrow b>0 \Rightarrow a^{2}=3 b^{2}+1 \leq 4 b^{2} \Rightarrow b^{\prime}=2 b-a \geq 0$; and $b>0 \Rightarrow 9 b^{2}>3 b^{2}+1=a^{2} \Rightarrow a-a^{\prime}=3 b-a>0$.

Thus, a solution $(a, b)$ in nonnegative integers with $a>1$ yields a new solution $\left(a^{\prime}, b^{\prime}\right)$ such that $a^{\prime}1$, we can repeat this process to obtain nonnegative integers $a^{\prime \prime}, b^{\prime \prime}$ such that $a^{\prime \prime}a^{\prime}>a^{\prime \prime}>\cdots$. But this sequence of nonnegative integers cannot decrease forever, so eventually we get to some $n$ for which $a^{(n)}=1$. The equation then implies $b^{(n)}=0$, so $a+b \sqrt{3}=(2+\sqrt{3})^{n}\left(a^{(n)}+b^{(n)} \sqrt{3}\right)=(2+\sqrt{3})^{n}$. Then, it follows that $(2-\sqrt{3})^{n}=a-b \sqrt{3}$. (This can be proven by comparing the binomial expansions of $(2+\sqrt{3})^{n}$ and $(2-\sqrt{3})^{n}$, or by a straightforward induction on $n$; note that the equation $a+b \sqrt{3}=(2+\sqrt{3})^{n}$ uniquely determines the integers $a$ and b.) Thus, we conclude that

$$
a=\frac{(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}}{2}, \quad b=\frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{2 \sqrt{3}} .
$$

If we had wanted all integer solutions to the equation $a^{2}-3 b^{2}=1$, rather than all nonnegative integer solutions, we simply note that the equation is invariant under switching signs of $a$ and/or $b$, so the set of all solutions is simply the set of solutions given above with possible sign changes thrown in.

Now we consider the given equation, $2 x^{2}-6 x y+3 y^{2}=-1$. This can be rewritten as $x^{2}-3(y-x)^{2}=1$. Thus, by letting $a=x, b=y-x$, we have an equivalence between our given equation and the Pell equation, so we can express our solutions to the former in terms of those of the latter: $x=a$ and $y=a+b$ gives

$$
x= \pm \frac{(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}}{2}, \quad y=x \pm \frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{2 \sqrt{3}}
$$

where $n$ is any nonnegative integer. It can be checked by computation that any such pair is a valid, integral solution, and the foregoing argument shows that all solutions are of this form.