The only solutions are $(2^k,2^k)$, $(p,1)$ (where p is prime) Let us suppose that there are solutions $(a,b)$ such that a and b are different numbers. and WLOG, $a>b$ Let the smallest prime number dividing a be denoted as $p_1$. $\phi(a)< (p_1^{e_1}-p_1^{e_1-1})p_2^{e_2}.....p_i{e_i}=({p-1})/p \times a$ $gcd(a,b)<a/p_1$ (as a and b are different) and $\phi(b)<b$ (by definition) and so $\phi (a) + \phi (b) + gcd (a, b)< ({p_1-1})/p_1 \times a + a/p_1 +b$ equality only holds when (a,b) is (p,1) and therefore there are no solutions where a and b are different except (p,1) Now let us suppose a and b are the same and let this number be n we must note here that if n is odd, then LHS is even and RHS is odd and therefore a clear contradiction Let $n= 2^k \times t$ where t is an odd number $n=2^{k-1} \times t$ will work too as $\phi(n/2)=1/2 \phi(n)$ when n is even and so both sides of the equations half by infinite dissent we will get $n=t$ working and this is a contradiction as t is odd therefore $2^k,2^k$ and (p,1) is the only solution and when we check this it works.

anthonyshinex wrote:

This is not true, $(p,1)$ is a solution for $p$ prime.

yes It is fixed sorry

Nice problem The solutions are $(x,y) \in \{(2^t,2^t);(1,p);(p,1)\}$, where $t \geq 1$, and $p$ is a prime number. Let $d=gcd(a,b)$. Then we have that $a=dx$ and that $b=dy$, for some $x$ and $y$. The the given equality is transformed into: $$d(x+y-1)=\varphi (d) \left(\varphi (x) + \varphi (y) \right)$$ Let's assume that $x,y$ are composite, also assume that $d \neq 1$ then we have that the following inequalities hold: $$d > \varphi (d) \; \; \text{and} \; \; x+y-1 > \varphi (x) + \varphi (y)$$This implies that we must look for the cases when those inequalities fail. So from here we have the following cases: The first case is when $d=1$ this implies that $x+y-1=\varphi (x) + \varphi(y)$, again assume that they are composite, this implies that $\frac{x}{2} \geq \varphi (x)$. This implies that: $$x+y-1 \leq \frac{x+y}{2} \implies x+y \leq 2$$This implies that $x=y=1$ which is a contradiction. Thus this means that we have that $x,y$ belong to the set of $\mathbb{P} \cup \{1\}$, where $\mathbb{P}$ is the set of primes. By a quick observation one notices that one of the is prime while the other one is equal to $1$. Thus we have a first set of solutions $(1,p)$ and $(p,1)$, where $p$ is prime. The second case is looking when $x+y-1 > \varphi (x) + \varphi(y)$ fails. By quick observation we have that $x=y=1$, this implies that $d=2\varphi(d)$, and this implies that $d=2^t$, for some $t$. By plugging in the information we have we get that $t \geq 1$

Smallest prime divisor.