If $ 3\mid p+1$, then $ ( x,y,n) =( 1,1,2( p+1) /3)$ satisfies the equation. We will show that if $ 3\nmid p+1$, the equation cannot hold. Suppose that $ 3\nmid p+1$ and $ x+1/x+p( y+1/y) =3n$ holds for some $ ( n,x,y) \in \mathbb{N\times Q}^{2}_{+}$. We can take $ ( a,b,c,d) \in \mathbb{N}^{4}$ such that $ \gcd( a,b) =\gcd( c,d) =1$ and $ ( x,y) =( a/b,c/d)$. So we have $ 3n=\frac{( ad+bc)( ac+pbd)}{abcd} =F\in \mathbb{N}$. In particular, we have $ abcd\mid ( ad+bc)( ac+pbd)$. We will use this divisibility relation repeatedly without notice. Let $ h=\gcd( b,d)$. Since $ h\mid b,d$ and $ \gcd( h,ac) =1$, we have $ h=1$. Since $ h=1$ and $ \gcd( a,b) =1$, we have $ b\mid c$. Similarly, we get $ d\mid a$. We can take $ ( s,t) \in \mathbb{N}^{2}$ such that $ a=ds$ and $ c=bt$. So we have $ F=\frac{( dds+bbt)( st+p)}{bdst}$. Let $ g=\gcd( s,t)$. We can easily get $ g\mid p$. So we consider the following two cases (1),(2). (1) $ g=1$ Since $ \gcd( s,bt) =1$, we have $ s\mid p$. Similarly, we get $ t\mid p$. So we consider the following three sub-cases $ ( \heartsuit )$,$ ( \diamondsuit )$. (1)-($ \heartsuit $) $ ( s,t) =( 1,1)$ In this case, $ F=\frac{\left( b^{2} +d^{2}\right)( p+1)}{bd}$. Since $ 3\nmid p+1$, we must have $ 3\mid b^{2} +d^{2}$. However, if $ 3\mid b^{2} +d^{2}$, then $ 3\mid b$ and $ 3\mid d$, which contradicts $ \gcd( b,d) =1$. (1)-($ \diamondsuit $) $ ( s,t) =( p,1)$ In this case, $ F=\frac{2( pdd +bb)}{bd}$. Since $ \gcd( b,dp) =1$, we have $ b\mid 2$ and $ d\mid 2$. In particular, $ \gcd( bd,3) =1$ becaues $ b,d\leq 2$. So $ pd^{2} +b^{2} \equiv p+1\bmod 3$, which is not divisbly by $ 3$. Thus we get a contradiction. (2) $ g=p$ We can take $ ( s,t) \in \mathbb{N}^{2}$ such that $ s=pu$,$ t=pv$, and $ \gcd( u,v) =1$. So we have $ F=\frac{( ddu+bbv)( p uv+1)}{bduv}$ Since $ \gcd( u,vb) =1$, we have $ u\mid 1$, which means $ u=1$. Similarly, we get $ v\mid 1$, which means $ v=1$. So we have $ F=\frac{( dd+bb)( p+1)}{bd}$. We can show that this is a contradiction in the same way as (1)-($ \heartsuit $). $ \blacksquare $