, 2020

First suppose that $a = 1$. Then $\varphi(1) = 1$. For all positive integers $b$ we have $\gcd(a, b) = 1$. Therefore in this case the equation is $1 + b = 1 + \varphi(b) + 1$, or equivalently, $\varphi(b) = b - 1$. This is equivalent to the statement that there exists a unique integer from $\{1, 2, \dots, b\}$; which then has to be $b$ itself (since unless $b = 1$, we have $\gcd(b, b) > 1$, but if $b = 1$ we have $\varphi(b) = b$). In other words, this is equivalent to $b$ being a prime number. Hence the solutions for $a = 1$ are precisely the pairs $(1, p)$ with $p$ a prime number. Similarly, the solutions for $b = 1$ are precisely the pairs $(p, 1)$ with $p$ a prime number.

Now assume that $a, b \ge 2$. As $\gcd(b, b) > 1$ we have $\varphi(b) \le b - 1$. Therefore
$$\gcd(a, b) = a + b - \varphi(a) - \varphi(b) \geq a - \varphi(a) + 1.$$
Let $p$ be the minimal prime divisor of $a$ (which exists as $a \ge 2$). Since for all multiples $tp \le a$ of $p$, we have $\gcd(tp, a) > 1$, it follows that $a - \varphi(a) \ge \frac{1}{p} \cdot a$. Therefore we have
$$\gcd(a, b) \geq a - \varphi(a) + 1 \geq \frac{a}{p} + 1.$$
The two largest divisors of $a$ are $a$ and $\frac{a}{p}$. Since $\gcd(a, b)$ is a divisor of $a$ that is at least $\frac{a}{p} + 1$, it must equal $a$. Hence $\gcd(a, b) = a$. In the same way we prove that $\gcd(a, b) = b$. So $a = b$.

The equation now is equivalent to $2a = 2\varphi(a) + a$, so also to $a = 2\varphi(a)$. Note that $2 \mid a$. Therefore write $a = 2^k \cdot m$ with $k \ge 1$ and $m$ odd. By a well-known property of the $\varphi$-function, we have $\varphi(a) = \varphi(2^k) \cdot \varphi(m) = 2^{k-1} \cdot \varphi(m)$, and the equation becomes $2^k \cdot m = 2 \cdot 2^{k-1} \cdot \varphi(m)$, or equivalently
$$m = \varphi(m).$$
Therefore $m = 1$, and $a = b = 2^k$. Indeed, the equation holds for all pairs $(2^k, 2^k)$ with $k \ge 1$.