In this post I evoke a variant of the equations showed in section D28 A reciprocal diophantine equation from [1], using particular values of the Euler's totient function $\varphi(n)$. I ask it from a recreational point of view to know how solve our equation $$\frac{1}{x}-\frac{1}{y}=\frac{1}{\varphi(xy)}\tag{1},$$ for integers $x,y\geq 1$. Our solutions will be denoted as pairs $(x,y)$.
Claim 0 (Preparation). Since $\frac{1}{\varphi(xy)}>0$ then $x<y$. Thus we consider integers $1\leq x<y$. And $(1)$ can be rewritten as $$(y-x)\cdot\prod_{\text{primes }p\mid xy}(p-1)=\prod_{\text{primes }p\mid xy}p\tag{1'}.$$
Computational facts. For integers $1\leq x,y\leq 500$ we can find the solutions $(x,y)=(2,4),$ $(3,6),(6,9),(9,12)$ and $(24,27)$. Our observations (I presume that these are the only solutions of our equation $(1)$) should be that the only solution $(x,y)$ with both $x$ and $y$ being even integers is $(2,4)$, and that seem that the other solutions have opposite parity (having $2$ and $3$ as only prime factors).
Question. I would like to know if our equation $(1)$ has a finite number of solution and if it is possible find a characterization of all the solutions of such equation. Thus I need to finish the reasoning to prove that previous are the solutions of our equation. Many thanks.
Here below I add the claims that I can deduce.
Claim 1. Let $(x,y)$ a solution such that $x$ and $y$ are both powers of two. Then $x=2$ and $y=4$.
Proof. Is $x=2^a$ and $y=2^b$, being $a$ and $b>1$ positive integers $\geq 1$. Then from $(1\text{'})$ one has that $2^a(2^{b-a}-1)=2$ implies $a=1$ and thus $b=a+1=2.\square$
Claim 2. A) There aren't solutions $(x,y)$ satisfying that $x\equiv y\equiv 0\text{ mod }2$ having or $x$ or $y$ an odd prime factor $P$ (that is, the only solution $(x,y)$ where both $x$ and $y$ are even integers is the previous $(2,4)$). B) From $(1\text{'})$ it is obvious that there aren't solutions $(x,y)$ such that both $x$ and $y$ are odd integers.
Proof. By contradition we assume $x\equiv y\equiv 0\text{ mod }2$ and that there exists an odd prime $P\mid x$ (the similar case, when there exists an odd prime dividing $y$, can be deduced), then from $(1\text{'})$ one deduces that $2||RHS$ of $(1\text{'})$ (this notation means that $RHS$ is even but $4\nmid RHS$) while that $4\mid LHS$, thus we get an absurd.$\square$
Claim 3. Thus with the exception of $(2,4)$ our solutions $(x,y)$ have opposite parity.
Final remarks.
A) The prime number $3$ is the only odd prime number such that $3-1=2$.
B) Being $P$ an odd prime factor dividing $x$ or dividing $y$, then it is obvious that the inventory of odd prime factors dividing $P-1$ will be the prime $3$ or (odd) prime numbers greater than $3$. The last case is absurd since if $3<Q$ is an odd prime factor dividing $P-1$, that is $Q\mid P-1$ then
subcase 1-then: $Q\mid RHS\text{ of }(1\text{'})$, and we get an absurd from here since $Q^2\mid RHS$ while $Q||RHS$.
subcase 2-then: this second case also an absurd, because the assumption of this second case is $Q\nmid RHS$, but our hypothesis says that $Q\mid P-1$ for some prime $P$ diving $xy$, thus $Q\mid LHS.$ $\square$
References:
[1] Richard K. Guy, Unsolved Problems in Number Theory, Volume I, Second Edition, Springer (1994).
