Solving for integral solutions in \(y^x=x^y\) for distinct \(x\) and \(y\) is equivalent to solving for integral solutions in \(y^{\frac{1}{y}}=x^{\frac{1}{x}}\) for distinct \(x\) and \(y\). Since \(\dfrac{d}{dx}\big(x^{\frac{1}{x}}\big)=\dfrac{1-\text{ln}(x)}{x^2}\times x^{\frac{1}{x}}\), we have a single maximum at \((e,e^{\frac{1}{e}})\) for \(x^{\frac{1}{x}}\) over positive \(x\). Then since \(x^{\frac{1}{x}}\rightarrow 0\) as \(x\rightarrow 0^+\), and \(x^{\frac{1}{x}}\) decreases asymptotically to \(1\) as \(x\rightarrow \infty \), all integral solutions must occur in the domain \([0,e]\); this leaves three possibilities for the value of \(x\): \(0\), \(1\), and \(2\). We have \(0^{\frac{1}{0}}\) to be undefined, and \(1^{\frac{1}{1}}\) to have no corresponding solution (since \(x^{\frac{1}{x}}\) only asymptotically approaches \(1\)), leaving only \(2^{\frac{1}{2}}\) to have the corresponding solution \(4^{\frac{1}{4}}\). Hence, the only solutions for distinct \(x\) and \(y\) over the positive integers are \((x,y)=(2,4),(4,2)\).

This question would have 2 solutions only when x is not equal to y.
In this question , it is not mention that x is not equal to y . So this question has infinite solutions
. When x=y (not mention in this question , so considering ) , then there are infinitely many solutions of (x, y) i.e. (1,1) ; (2,2) ; (3,3) ; (4,4) .............

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TopNewestSolving for integral solutions in \(y^x=x^y\) for distinct \(x\) and \(y\) is equivalent to solving for integral solutions in \(y^{\frac{1}{y}}=x^{\frac{1}{x}}\) for distinct \(x\) and \(y\). Since \(\dfrac{d}{dx}\big(x^{\frac{1}{x}}\big)=\dfrac{1-\text{ln}(x)}{x^2}\times x^{\frac{1}{x}}\), we have a single maximum at \((e,e^{\frac{1}{e}})\) for \(x^{\frac{1}{x}}\) over positive \(x\). Then since \(x^{\frac{1}{x}}\rightarrow 0\) as \(x\rightarrow 0^+\), and \(x^{\frac{1}{x}}\) decreases asymptotically to \(1\) as \(x\rightarrow \infty \), all integral solutions must occur in the domain \([0,e]\); this leaves three possibilities for the value of \(x\): \(0\), \(1\), and \(2\). We have \(0^{\frac{1}{0}}\) to be undefined, and \(1^{\frac{1}{1}}\) to have no corresponding solution (since \(x^{\frac{1}{x}}\) only asymptotically approaches \(1\)), leaving only \(2^{\frac{1}{2}}\) to have the corresponding solution \(4^{\frac{1}{4}}\). Hence, the only solutions for distinct \(x\) and \(y\) over the positive integers are \((x,y)=(2,4),(4,2)\).

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Do you mean it has 3 solutions?

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I meant writing 3.

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This question would have 2 solutions only when x is not equal to y. In this question , it is not mention that x is not equal to y . So this question has infinite solutions . When x=y (not mention in this question , so considering ) , then there are infinitely many solutions of (x, y) i.e. (1,1) ; (2,2) ; (3,3) ; (4,4) .............

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I am satisfied with (2, 4) and (4, 2). Where (2, 4) is an obvious solution. (0,0) isn't a solution.

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I meant writing (1, 1). I was way to hurried.

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