Can anybody help me with this equation?
Solve in $\mathbb{N}$:
$$3x^2 - 7y^2 +1=0$$
One solution is the pair $(3,2)$, and i think this is the only pair of positive integers that can be a solution. Any idea?
Answer
There are infinitely many solutions in positive integers. $7y^2-3x^2=1$ is an example of a "Pell equation", and there are standard methods for finding solutions to Pell equations.
For example, the fact that $(x,y)=(2,3)$ is a solution to $7y^2-3x^2=1$ is equivalent to noting that $(2\sqrt7+3\sqrt3)(2\sqrt7-3\sqrt3)=1$. The fundamental unit in $\Bbb Q(\sqrt{21})$ is $55+12\sqrt{21}$; in particular, $(55+12\sqrt{21})(55-12\sqrt{21})=1$. Consequently, if we calculate $(2\sqrt7+3\sqrt3)(55+12\sqrt{21}) = 218 \sqrt{7}+333 \sqrt{3}$, it follows that $(218 \sqrt{7}+333 \sqrt{3})(218 \sqrt{7}-333 \sqrt{3})=1$, or $7\cdot218^2 - 3\cdot333^2=1$.
You can get infinitely many solutions $(x_n,y_n)$ to $7y^2 - 3x^2=1$ by expanding $(2\sqrt7+3\sqrt3)(55+12\sqrt{21})^n = y_n \sqrt{7}+x_n \sqrt{3}$. Your solution corresponds to $n=0$, while the previous paragraph is $n=1$; for example, $n=2$ yields $x_2=36627$ and $y_2=23978$.
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