1. Find all pairs of natural numbers $a$ and $b$ for which $2 a^{b}=a b+3$.

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IV/1. 1. way. Since $b$ is a natural number, $a$ divides $2 a^{b}$ and $a b$, so $a$ also divides 3. Since 3 is a prime number, $a=1$ or $a=3$. If $a=1$, we get the equation $2=b+3$, which has no solutions in natural numbers. Therefore, $a=3$, from which it follows that $2 \cdot 3^{b}=3 b+3$ or $2 \cdot 3^{b-1}=b+1$. One solution to this equation is $b=1$. We will prove that there are no other solutions in natural numbers. We will do this by proving by induction on $b$ that $2 \cdot 3^{b-1}>b+1$ for all $b \geq 2$. For $b=2$, this is true, since $2 \cdot 3^{1}=6>3=2+1$. Assume that $2 \cdot 3^{b-1}>b+1$ for some $b \geq 2$. Then $2 \cdot 3^{b}=3 \cdot 2 \cdot 3^{b-1}>3(b+1)>b+2$. The induction is thus complete. Therefore, it is indeed true that $2 \cdot 3^{b-1}>b+1$ for all $b \geq 2$, which means that $b=1$ is the only solution to the equation $2 \cdot 3^{b-1}=b+1$ in natural numbers.

The given equation is solved only by the pair $a=3$ and $b=1$.

2. way. As in the first solution, we conclude that $a=3$ and $2 \cdot 3^{b}=3 b+3$. Consider the functions $f(x)=3 x+3$ and $g(x)=2 \cdot 3^{x}$. The real solutions to the equation $2 \cdot 3^{b}=3 b+3$ are the points where the graphs of the functions $f$ and $g$ intersect. Function $f$ is a linear function, and function $g$ is a scalar multiple of an exponential function. The graphs of such two functions intersect at most twice. Since $f(1)=6=g(1)$, one intersection point is at $x=1$. The second intersection point lies in the interval between -1 and 0, since $f(-1)=02=g(0)$. The equation $2 \cdot 3^{b}=3 b+3$ thus has two real solutions, but only $b=1$ is a natural number.

Therefore, the given equation is solved only by the pair $a=3$ and $b=1$.

Observation that $a$ divides 3 or that $a$ and $b$ are odd numbers

1 point

Writing that for $a=3$, we get the equation $2 \cdot 3^{b-1}=b+1$ 1 point

Solution $a=3$ and $b=1$ 1 point

Justification that this is the only solution - justification that $a \leq 3 \ldots \ldots \ldots \ldots \ldots .2$ points Justification that this is the only solution - that for $a=3, b=1$ is the only solution to the equation $2 \cdot 3^{b-1}=b+1$ 2 points