<s>[INST] Find all values for x and y, where x and y are prime. So the question is as follows:

Find all values for $x$ and $y,$ where $x$ and $y$ are prime, and $3 (x+1) (y+1) \equiv 0 \pmod {xy}.$

I tried this and figured out that (2,3) is one of the solution. But I am confused how to prove that's the unique solution!
Please help me.
Thank you
 [/INST] We have $3(x+1)(y+1) = xyk$, $k\in\mathbb{Z}$, so either $3\mid k$ or $3 \mid xy$.

*

*If $3 \mid k$, then we have $k = 3n$, $n\in\mathbb{Z}$, so $$(x+1)(y+1) = xyn.$$
We know $x+1 \nmid x \implies x+1 \mid yn$. Then, we have $(x+1)m = yn$, so $y+1 = xm \implies x \mid y+1$. With the same logic, we get $y\mid x+1$. If $y$ is an odd prime, $x = 2$ from the first relation and $y = 3$ from the second. If $y=2$, then $x=3$, so we get two solutions, $(2,3)$ and $(3,2)$.


*If $3 \mid xy$, then either $x=3$ or $y=3$, since both are prime. WLOG, let $x=3$. Then, $$4(y+1) = yk.$$ Since $y+1 \nmid y$, $y+1 \mid k$. In addition, we know $4 \mid yk$. If $4 \mid k$, then $4(y+1)\mid k \implies y = 1$ which leads to a contradiction. Thus, we must have $2\mid y$ and $2\mid k$. This leads to the solutions $(3,2)$ and $(2,3)$.
Hence the only values are $(2,3)$ and $(3,2)$.
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