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Here $\varphi(n)$ denotes the Euler's totient function. I've deduced a family of prime numbers $$(x=p_1,y=p_2)$$ that solve an equation involving the Euler's totient function. These primes are the prime numbers that appear in the Rassias' conjecture, see this Wikipedia, since I've created this equation on assumption of this conjecture with this purpose. Now I try to find different sets of positive integers being solution of the equation that I present.

Claim. Let $k\geq 1$ a fixed integer. On assumption of the Rassias' conjecture one has that there exist infinitely many solutions of the equation $$\varphi\left(\left(x+y+1\right)^k\right)=\left(1-\frac{1}{x}\right)(y+1)\left(x+y+1\right)^{k-1}\tag{1}$$ with $x<y$ prime numbers .

Question. Can you find a different infinite family of integer solutions, $1<u<v$ not being both prime numbers $$\varphi\left(\left(u+v+1\right)^\kappa\right)=\left(1-\frac{1}{u}\right)(v+1)\left(u+v+1\right)^{\kappa-1},$$ for a suitable $\kappa>1$ fixed integer? Many thanks.

If it is very difficult but you've computational evidence, feel free to add this evidence of the existence of such set of solutions.

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An infinite family of solutions is given by: $$(u,v)=(p,pq-p-1)$$ where $p$ and $q$ are different primes numbers.


Proof \begin{align*} \varphi((u+v+1)^\kappa) &=\varphi(p^\kappa q^\kappa)\\ &=\varphi(p^\kappa)\varphi(q^\kappa)\\ &=(p-1)(q-1)(pq)^{\kappa-1}\\ &=(u-1)\cdot\frac{v+1}{u}(v+u+1)^{\kappa-1}\\ &=\left(1-\frac1u\right)(v+1)(v+u+1)^{\kappa-1} \end{align*}

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  • $\begingroup$ Now I'm stuck with the fourth identiy, but tomorrow I check it. Many thanks. Also I am going to wait if there are more answers, but I should accept yours as right $\endgroup$ – user243301 Jan 18 '18 at 21:15

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