### 7.4. Consider the expression

$$
E(a, b)=\frac{3}{a+2}+\frac{7}{2 b+3}
$$

Determine all non-zero natural numbers $a$ and $b$, for which the value of the expression $E(a, b)$ is a natural number
## Solution:

Since $0<\frac{3}{a+2} \leq 1$ and $0<\frac{7}{2 b+3}<2 \Rightarrow E(a, b) \in(0 ; 3)$. Since $E(a, b) \in \mathbb{N} \Rightarrow E(a, b)$ $\in\{1,2\}$.

Case 1:

Let $E(a, b)=1 \Leftrightarrow \frac{3}{a+2}+\frac{7}{2 b+3}=1 \Rightarrow \frac{3}{a+2}=\frac{2 b-4}{2 b+3} \Rightarrow \frac{a+2}{3}=\frac{2 b+3}{2 b-4} \Rightarrow a$

$$
=1+\frac{21}{2 b-4}
$$

Since $a \in \mathbb{N}^{*} \Rightarrow \frac{21}{2 b-4}$

$\in \mathbb{N}^{*}$ - impossible because $2 b-4$ is an even number and 21 is odd.

## Case 2:

$$
\text{Let } \begin{aligned}
E(a, b)=2 & \Leftrightarrow \frac{3}{a+2}+\frac{7}{2 b+3}=2 \Rightarrow \frac{3}{a+2}=\frac{4 b-1}{2 b+3} \Rightarrow \frac{a+2}{3}=\frac{2 b+3}{4 b-1} \Rightarrow a \\
& =\frac{11-2 b}{4 b-1}
\end{aligned}
$$

Since $a \in \mathbb{N}^{*} \Rightarrow 11-2 b \geq 4 b-1 \Rightarrow b \leq 2$. Since $b \in \mathbb{N}^{*} \Rightarrow b \in\{1,2\}$. For $b=1 \Rightarrow$ $a=3$,

and for $b=2 \Rightarrow a=1$.

Answer: $(a, b) \in\{(1,2) ;(3,1)\}$.