Follow-up 'Diophantine' question: extrapolating to the general case

Question

I came across a problem involving a certain Diophantine equation a few days ago. I learnt quite a few extremely helpful things about them on this thread here, which I started: A model that can be followed when solving Diophantine equations - ideas? -

Feel free to read it for context for what follows:

Whilst toiling with the concept of Diophantine equations I seemed to notice a surprising pattern.

Any equation of the form:

$$\frac{1}{x}+\frac{2}{y}=\frac{3}{p}$$

(Where x and y are positive integers and p is a prime greater than or equal to 5)

always seems to have exactly 3 solutions.

Is it possible to prove this initial hypothesis/conjecture? Or have I thought incorrectly and can you disprove me? Looking forward to seeing what the responses yield!

Answer 1

$$ 3xy - 2px - py = 0$$

$$ 9xy - 6px - 3py = 0$$

$$ 9xy - 6px - 3py + 2p^2 = 2p^2$$

$$ (3x-p)(3y-2p) = 2p^2$$

\begin{array}{|c|c|c|c|c|c|} \hline &\color{red}{(3x-p)}\color{blue}{(3y-2p)} & x & y\\ \hline A&\color{red}{1}\cdot\color{blue}{2p^2} & (p+1)/3 & 2p(p+1)/3 & \color{green}{?}\\ \hline B&\color{red}{2p^2}\cdot\color{blue}{1} & p(2p+1)/3 & (2p+1)/3 & \color{green}{?}\\ \hline C&\color{red}{2}\cdot\color{blue}{p^2} & (p+2)/3 & p(p+2)/3 &\color{green}{?}\\ \hline D&\color{red}{p^2}\cdot\color{blue}{2} & p(p+1)/3 & 2(p+1)/3 &\color{green}{?}\\ \hline E&\color{red}{p}\cdot\color{blue}{2p} & 2p/3 & 4p/3 &\color{red}{✗} \\ \hline F&\color{red}{2p}\cdot\color{blue}{p} & p & p &\color{green}{\checkmark} \\ \hline \end{array}

From the table we come across two cases :

• When $p+1 \equiv 0 \pmod 3$, $A$, $D$ and $F$ each produce one solution.

E.g. for $p=11$, $$ {\dfrac{1}{4}+\dfrac{2}{88}=\dfrac{3}{11},\dfrac{1}{44}+\dfrac{2}{8}=\dfrac{3}{11},\dfrac{1}{11}+\dfrac{2}{11}=\dfrac{3}{11}}$$

• When $p+2 \equiv 0 \pmod 3$, $B$, $C$ and $F$ each produce one solution.

E.g. for $p=7$, $$ {\dfrac{1}{35} +\dfrac{2}{5}=\dfrac{3}{7},{\dfrac{1}{3}+\dfrac{2}{21}=\dfrac{3}{7}},\dfrac{1}{7}+\dfrac{2}{7}=\dfrac{3}{7}}$$

Note that $2p+1 \equiv 0 \equiv 2p+1+3 \equiv p+2 \pmod 3$.

Thus the conjecture that $1/x + 2/y = 3/p$ has exactly three solutions in positive integers $x$, $y$ for prime $p \ge 5$ is correct!

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Thanks to @Servaes for pointing out the critical mistake! :)