Find all prime numbers $p$, for which there exist $x, y \in \mathbb{Q}^+$ and $n \in \mathbb{N}$, satisfying $x+y+\frac{p}{x}+\frac{p}{y}=3n$.
Problem
Source: VIII International Festival of Young Mathematicians Sozopol 2017, Theme for 10-12 grade
Tags: algebra, number theory, prime numbers
kaede
21.01.2020 10:34
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 $