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The equation is: $$ 2^x3^{x-1}=y\cdot3^{x-1}\cdot2+z\cdot2^{x-1}$$ for natural numbers.

I’ve tried to divide this expression or try various substitutions, but nothing is working.

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    $\begingroup$ Can you find any solutions? Do your natural numbers include zero? $\endgroup$ Commented Oct 21, 2021 at 0:54
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    $\begingroup$ Note that $z$ must have a factor $3^{x-1}$ so divide that out. Now $y$ needs a factor $2^{x-2}$ $\endgroup$ Commented Oct 21, 2021 at 0:58
  • $\begingroup$ I’ve guessed some solutions, e.g.: (2, 2, 9). My definition of natural numbers doesn’t include zero. $\endgroup$
    – Dodomol
    Commented Oct 21, 2021 at 1:06

2 Answers 2

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I'm gonna assume natural numbers don't include $0$ for simplicity.

If $x=1$, then $2=2y+z$, which has no solution, as $2y+z\ge 2 + 1 = 3$.

Assume $x\ge2$, then $3^{x-1} | z\cdot 2^{x-1}$, therefore $3^{x-1} | z$, so $z=3^{x-1}c$ for some natural number $c$. Similarly, $2^{x-1}|2y$, $y=2^{x-2}b$, therefore the equation becomes $4=2b+2c$ which has a unique solution $b=c=1$. Therefore, the general solution is just $(x, 2^{x-2}, 3^{x-1})$.

So the solutions are $\{(x, 2^{x-2}, 3^{x-1}) \mid x\in\mathbb N, x\ge 2\}$

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$$ \begin{array} {} 2^x3^{x-1} &= y \cdot 3^{x-1}\cdot 2 + z \cdot 2^{x-1} & \text {use $w$ for $x-1$} \\ 2 \cdot 2^w 3^w &= y \cdot 3^w\cdot 2 + z \cdot 2^w \\ 2 \cdot 3^w \cdot(2^w-y) &= z \cdot 2^w &\implies z=z_1 3^w\\ 2 \cdot 3^w \cdot(2^w-y) &= z_1 \cdot 3^w \cdot 2^w \\ 2 \cdot(2^w-y) &= z_1 \cdot 2^w \\ 2^w \cdot (2-z_1)&= 2 \cdot y &\implies y=y_1 \cdot 2^{w-1} \\ 2= y_1 + z_1 & \land y,z \in \mathbb N &\implies y_1=z_1=1 \\ \implies \\ (4y,3z)&=(2^x,3^x) & \text{ insert & check! } \\ 2^x3^{x-1} &= 2^{x-2} \cdot 3^{x-1}\cdot 2 + 3^{x-1} \cdot 2^{x-1} \\ 2^x3^{x-1} &= 2^{x-1} \cdot 3^{x-1} + 3^{x-1} \cdot 2^{x-1} \\ 2^x &= 2^{x-1} + 2^{x-1} = 2^x & \implies \text{result is correct!} \\ \end{array} $$

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