The number of ordered pairs (x, y) of positiv | vedclass

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(A) Given the equation $2^x + 3^y = 5^{xy}$ for positive integers $x, y \in \mathbb{Z}^+$.If $x = 1$ and $y = 1$,then $2^1 + 3^1 = 2 + 3 = 5$ and $5^{

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(A) Given the equation $2^x + 3^y = 5^{xy}$ for positive integers $x, y \in \mathbb{Z}^+$.If $x = 1$ and $y = 1$,then $2^1 + 3^1 = 2 + 3 = 5$ and $5^{1 	imes 1} = 5^1 = 5$. Thus,$(1, 1)$ is a solution.If $x > 1$ or $y > 1$,we can divide the equation by $5^{xy}$ to get $\frac{2^x}{5^{xy}} + \frac{3^y}{5^{xy}} = 1$.This can be rewritten as $\left(\frac{2}{5^y}ight)^x + \left(\frac{3}{5^x}ight)^y = 1$.For $x, y \ge 1$,the terms $\frac{2^x}{5^{xy}}$ and $\frac{3^y}{5^{xy}}$ decrease rapidly as $x$ or $y$ increase.Specifically,for $x, y \ge 1$,$2^x + 3^y Therefore,the only ordered pair $(x, y)$ that satisfies the equation is $(1, 1)$.

The number of ordered pairs $(x, y)$ of positive integers satisfying $2^x + 3^y = 5^{xy}$ is

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