, 2015

Since $b$ is a positive integer, $a$ divides $2a^b$ and $ab$, and hence also $3$. Since $3$ is a prime number, we have $a = 1$ or $a = 3$.

If $a = 1$ we get equation $2 = b + 3$, which does not have solutions in positive integers.

Therefore $a = 3$ and we get equation $2 \cdot 3^b = 3b + 3$ which gives $2 \cdot 3^{b-1} = b + 1$.

One solution of this equation is $b = 1$.

Let's prove that there are no other solutions in positive integers. We shall use induction with respect to $b$ and prove that $2 \cdot 3^{b-1} > b + 1$ for all $b \ge 2$.

For $b = 2$ this is true since $2 \cdot 3^1 = 6 > 3 = 2 + 1$.

Suppose that $2 \cdot 3^{b-1} > b + 1$ for some $b \ge 2$. Then $2 \cdot 3^b = 3 \cdot 2 \cdot 3^{b-1} > 3(b+1) > b + 2$, and induction is finished.

Thus we indeed have that $2 \cdot 3^{b-1} > b + 1$ for all $b \ge 2$, which means that $b = 1$ is the only solution of the equation $2 \cdot 3^{b-1} = b + 1$ in positive integers.

The only solution to the given equation is the pair $a = 3$ and $b = 1$.

As in Solution 1 we deduce that $a = 3$ and $2 \cdot 3^b = 3b + 3$. Let's take a look at the functions $f(x) = 3x + 3$ and $g(x) = 2 \cdot 3^x$. Real solutions of the equation $2 \cdot 3^b = 3b + 3$ are those points at which the graphs of functions $f$ and $g$ intersect. The function $f$ is a linear function, the function $g$ is a scalar multiple of the exponent function. The graphs of such two functions intersect at most twice. Since $f(1) = 6 = g(1)$ one intersection is at $x = 1$. The second intersection lies on the interval between $-1$ and $0$ since $f(-1) = 0 < \frac{2}{3} = g(-1)$ and $f(0) = 3 > 2 = g(0)$. The equation $2 \cdot 3^b = 3b + 3$ thus has two real solutions, only one of which is a positive integer.

The only solution to the given equation is the pair $a = 3$ and $b = 1$.