Problem 4. Find the natural numbers $x, y, z$ that satisfy the equation:

$$
2^{x}+3 \cdot 11^{y}=7^{z}
$$
Solution. If $x=0,1+3 \cdot 11^{y}=7^{z}$, impossible due to parity.

Since $7^{z}=\mathcal{M} 3+1,3 \cdot 11^{y}=\mathcal{M} 3$ we deduce that $x$ is even. ................ 1p

If $z$ is odd then $x=2$, otherwise, if by contradiction $x \geq 4$ we obtain $2^{x}=\mathcal{M} 8$, and depending on the parity of $y, 3 \cdot 11^{y}=\mathcal{M} 8+3$ or $3 \cdot 11^{y}=\mathcal{M} 8+1$, impossible since $7^{z}=\mathcal{M} 8-1$.

If $x=2$ then $4+3 \cdot 11^{y}=7^{z} \Rightarrow y=0$, otherwise if $y \geq 1 \Rightarrow 7^{z} \equiv 4(\bmod 11)$, impossible for $z$ odd, since $7^{10 k} \equiv 1(\bmod 11), 7^{10 k+1} \equiv 7(\bmod 11), 7^{10 k+2} \equiv 5$ $(\bmod 11), 7^{10 k+3} \equiv 2(\bmod 11), 7^{10 k+4} \equiv 3(\bmod 11), 7^{10 k+5} \equiv 10(\bmod 11), 7^{10 k+6} \equiv 4$ $(\bmod 11), 7^{10 k+7} \equiv 6(\bmod 11), 7^{10 k+8} \equiv 9(\bmod 11), 7^{10 k+9} \equiv 8(\bmod 11)$.

![](https://cdn.mathpix.com/cropped/2024_06_07_64e54d985065c186c12ag-4.jpg?height=52&width=1537&top_left_y=1058&top_left_x=283)

If $z$ is even, $z=2 z_{1}, z_{1} \geq 1$ and $x=4 x_{1}, x_{1} \geq 1$ we obtain $3 \cdot 11^{y}=7^{2 z_{1}}-2^{4 x_{1}}=$ $\left(7^{z_{1}}-2^{2 x_{1}}\right) \cdot\left(7^{z_{1}}+2^{2 x_{1}}\right)$.

Since $\left(7^{z_{1}}-2^{2 x_{1}}, 7^{z_{1}}+2^{2 x_{1}}\right)=1$ and $7^{z_{1}}+2^{2 x_{1}}=\mathcal{M} 3+2$, we obtain the system of equations $7^{z_{1}}-2^{2 x_{1}}=3,7^{z_{1}}+2^{2 x_{1}}=11^{y}$. If $z_{1}$ were even, $z_{1}=2 z_{2}$, from the first equation we would obtain $\left(7^{z_{2}}-2^{x_{1}}\right) \cdot\left(7^{z_{2}}+2^{x_{1}}\right)=3$, which is impossible. We deduce that $z_{1}$ is odd and passing to the second equation, $2^{2 x_{1}}=11^{y}-7^{z_{1}}$ is either $\mathcal{M} 8+2$ or $\mathcal{M} 8+4$. We conclude that $x_{1}=1, z_{1}=1$ and from here the solution $x=4, y=1, z=2$.

If $z$ is even, $z=2 z_{1}, z_{1} \geq 1$ and $x=4 x_{1}+2, x_{1} \geq 0$, we obtain the analogous decomposition, $3 \cdot 11^{y}=7^{2 z_{1}}-2^{4 x_{1}+2}=\left(7^{z_{1}}-2^{2 x_{1}+1}\right) \cdot\left(7^{z_{1}}+2^{2 x_{1}+1}\right)$, and since $7^{z_{1}}-2^{2 x_{1}+1}=\mathcal{M} 3+2$ and $\left(7^{z_{1}}-2^{2 x_{1}+1}, 7^{z_{1}}+2^{2 x_{1}+1}\right)=1$ we obtain the system of equations $7^{z_{1}}-2^{2 x_{1}+1}=11^{y}, 7^{z_{1}}+2^{2 x_{1}+1}=3$ which clearly has no solution.