I'm interested in finding positive integers which satisfy an equation.
I've been thinking about the following equation:
$$x^2-10y^2=1\ \ \ \ \ \ \cdots(\star).$$
Then, I've just got the following (let's call this theorem):
Theorem: If $(x,y)$ satisfies $(\star)$, then $(20y^2+1,2xy)$ also satisfies $(\star)$.
Proof: Letting $x=10n+1$, we get $2n(5n+1)=y^2$. Hence, let's consider the case in which both $2n$ and $5n+1$ are square numbers. Then, we can represent $n=2k^2$, so we get $5n+1=10k^2+1$. Hence, $y=k$ is sufficient because of $(\star)$.
Then, letting $n=2y^2$, then we get
$$2n(5n+1)=2\times2y^2\times(10y^2+1)=(2xy)^2\ \Rightarrow\ (20y^2+1)^2-10(2xy)^2=1$$
Now the proof is completed.
It's easy to get $(x,y)=(19,6)$, so by using this theorem, we know we can get an infinite number of sets $(x,y)$ as the following:
$$(19,6), (721,228), (1039681,328776),\cdots$$
By the way, by using computer, I found that $(x,y)=(27379,8658)$ also satisfies $(\star)$ though this set cannot be got from the theorem above.
Then, here is my question.
Question: How can we get all positive-integers sets $(x,y)$ which satisfies $(\star)$ ?
I've tried, but I'm facing difficulty. Any help would be appriciated.
Best Answer
Note that if $a^2-10b^2=1$ we have $(a+\sqrt{10}b)(a-\sqrt{10}b)=1$ and also therefore that $$(a+\sqrt{10}b)^r(a-\sqrt{10}b)^r=1$$ So that $(a+\sqrt{10}b)^r=A+\sqrt{10}B$ generates a solution $(A,B)$
Also if we know that $(a,b)$ is a solution, and $(c,d)$ is a solution then
$$(a+\sqrt{10}b)(a-\sqrt{10}b)(c+\sqrt{10}d)(c-\sqrt{10}d)=1$$
and we can take $(a+\sqrt{10}b)(c+\sqrt{10}d)=(ac+10bd)+(ad+bc)\sqrt{10}$ which yields the new solution $(ac+10bd, ad+bc)$ - and we can always use the smallest solution $(19,6)$ for $(a,b)$