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Problem: Find all positive integer solutions to the Diophantine equation: $$3x^{2}- 5y^{2}= 7$$

Attempted approach:
Rewriting the equation: $$x^{2}= \frac{5y^{2}+ 7}{3}$$ We want the right-hand side to be an integer and a perfect square, so we first require that $5y^{2}+ 7\equiv 0\pmod 3$.
Since $5\equiv 2\pmod 3$, this becomes: $$2y^{2}+ 7\equiv 0\pmod 3\Rightarrow 2y^{2}\equiv -7\equiv 2\pmod 3\Rightarrow y^{2}\equiv 1\pmod 3$$ This gives a necessary condition for $y$, but I am unsure how to proceed from here.
I also considered multiplying the entire equation by a constant $k$, leading to: $$\left ( 3kx \right )^{2}- \left ( 5ky \right )^{2}= 7k$$ but couldn't find a helpful choice of $k$ that makes both sides squares.

Question: Could you help continue from here or suggest a better method to find all positive integer solutions to the original equation?

Dang Dang
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1234567890
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1 Answers1

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We can easily verify that $(x,y)=(a,b)$ is solution of $3x^2-5y^2=7$ if and only if $(X,Y)=(3a,b)$ is solution of $X^2-15Y^2=21$. $X^2-15Y^2=21$ is generalized Pell's equation and we can use similar method to solve Pell's equation.

$(X,Y)=(6,\pm1)$ is solution of $X^2-15Y^2=21$ with minimal $X$, and we can easily verify that fundamental solution of Pell's resolvent $u^2-15v^2=1$ is $(u,v)=(4,1)$. So all solutions of $X^2-15Y^2=21$ $(X_n^+,Y_n^+),(X_n^-,Y_n^-)$ can be obtained with $X_n^++Y_n^+\sqrt{15}=(6+\sqrt{15})(4+\sqrt{15})^n\quad(n\ge 0)$ and $X_n^-+Y_n^-\sqrt{15}=(6-\sqrt{15})(4+\sqrt{15})^n\quad(n>0)$.

First some solutions of $X^2-15Y^2=21$ obtained by this method is $(X_n^+,Y_n^+)=(6,1),(39,10),(306,79),(2409,622),(18966,4897)$ and $(X_n^-,Y_n^-)=(9,2),(66,17),(519,134),(4086,1055),(32169,8306)$. Using this, we can obtain first ten solution of $3x^2-5y^2-7$ as follows.

\begin{align*} (x,y)= & (2,1),(3,2),(13,10),(22,17),(102,79),\\ & (173,134),(803,622),(1362,1055),(6322,4897),(10723,8306) \end{align*}

Myungheon Lee
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  • So is there a way to find all of the positive integer solutions, for example like a generalised formula? – 1234567890 May 03 '25 at 09:42
  • @1234567890 if you label Myungheon's pairs, in order, as $(x_n, y_n)$ there are recursions for each letter: $x_{n+4} = 8 x_{n+2} - x_n,$ then $y_{n+4} = 8 y_{n+2} - y_n.$ Thus there is a formula for odd $n$ in terms of $(4+ \sqrt{15})^n $ and its reciprocal, slightly different formula for even $n.$ Each quite similar to Binet's formula for Fibonacci numbers – Will Jagy May 03 '25 at 18:36
  • The rules still apply if we begin the list of pairs as $(22, -17), (13, -10), (3,-2), (2,-1), (2,1),(3,2), (13,10), (22,17), (102,79), (173,134),...$ – Will Jagy May 03 '25 at 19:03
  • and the mapping (positive determinant) that preserves the value of $3 x^3 - 5 y^2$ is $$ (x,y) \mapsto (4x+5y, 3x+4y) $$ Check, easy to see that the mapping takes a pair not to the immediate next larger pair, but to the second pair over, as $(2,1) \mapsto(13,10)$ – Will Jagy May 03 '25 at 19:27