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 $Q\mid LHS.$ $\square$


[1] Richard K. Guy, Unsolved Problems in Number Theory, Volume I, Second Edition, Springer (1994).

  • $\begingroup$ I hope that the grammar of my claims was right. I am looking a justification of why our solutions are those showed in Computational facts. Thus I need to finish such reasoning from my claims, if it is feasible. $\endgroup$
    – user243301
    Commented May 26, 2018 at 15:34
  • 1
    $\begingroup$ No further solution for $1\le x\le y\le 10^4$ $\endgroup$
    – Peter
    Commented May 26, 2018 at 16:24
  • $\begingroup$ Many thanks for your attention (yours and all users) and help for such computational evidence. Feel free to study different generalizations of my equation $(1)$ involving the Euler's totient function @Peter $\endgroup$
    – user243301
    Commented May 26, 2018 at 16:30
  • $\begingroup$ I worked out that the list of the solutions is complete. The first step is to show that $xy$ must be of the form $2^m\cdot 3^n$ with $m>0$ and $n\ge 0$. Could you finish this step ? $\endgroup$
    – Peter
    Commented May 26, 2018 at 17:47
  • $\begingroup$ It is the question that I tried to solve with my claims @Peter , thus if you want add an answer. $\endgroup$
    – user243301
    Commented May 26, 2018 at 17:49

1 Answer 1


Hint : The given equation is equivalent to $$\frac{xy}{\phi(xy)}=y-x$$

So with $m:=xy$ we have $\phi(m)|m$ . If $p_1,\cdots, p_n$ are the prime factors of $n$, we have $$\frac{m}{\phi(m)}=\frac{p_1}{p_1-1}\cdots \frac{p_n}{p_n-1}$$

Now, show that more than one odd prime factor of $n$ is impossible, then show, if there is another odd prime factor, it must be $3$. The proof can be finished with Catalan's conjecture (which is proven)

  • 1
    $\begingroup$ Many thaks for your answer. $\endgroup$
    – user243301
    Commented May 26, 2018 at 17:56
  • $\begingroup$ I know how to prove from your equation the fact that there aren't two odd prime numbers dividing $m=xy$ (we are working the case of $x$ and $y$ having opposite parity) since $\nu_2(m/\varphi(m))\geq 0$ and if we assume two distinct odd prime numbers dividing $m$ then $\nu_2(m/\varphi(m))+\nu_2\left(\prod_{\text{primes }p\nmid m}(p-1)\right)\geq 2$ while $\nu_2\left(\prod_{\text{primes }p\nmid m}(p-1)\right)=1$. When I've introduced in the equation $(1\text{'})$ (the cases for our $x$ and $y$ of different parity) $xy=3^a2^b$ I can get equations that should be solved using Catalan's conjecture. $\endgroup$
    – user243301
    Commented May 26, 2018 at 20:51
  • $\begingroup$ But still I am thinking in the details of the argument (the only odd prime number dividing $m=xy$ will be $3$). I add these words as apologize of why I didn't accept the answer (I need to think more about it). Many thanks again for your answer. $\endgroup$
    – user243301
    Commented May 26, 2018 at 20:54

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