# If $x^y + y^x = 84$ and $x>3$, find the value of $x$ and $y$.

I see that taking log on both sides is not an option since the LHS consists of a sum and not a product.

I tried differentiating it but without any boundary conditions I fail to see any possible solutions.

Can someone guide me on how to tackle this question?

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$x,y$ integers, or any reals? –  gt6989b Mar 5 '13 at 17:02
not specified. I'd assume them to be integers. –  mathboggled Mar 5 '13 at 17:05
By the way, the above expression is not a polynomial. –  Thomas Andrews Mar 5 '13 at 17:18

Hint: There aren't many choices for $y$. It can only be $1,2,$ or $3,$ because if it is $4$ or greater $x^y \gt 84$ So try each of them in turn and see what you find.

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That makes sense. Thanks! –  mathboggled Mar 5 '13 at 17:17
That's not quite true. If $x$ is 1 $y$ can be as large as $83$ (which actually gives a solution). –  Elmar Zander Mar 5 '13 at 17:41
@ElmarZander: but we were given $x \gt 3$. Otherwise, you are correct –  Ross Millikan Mar 5 '13 at 18:06
@RossMillikan Oops, you're right. I overlooked that. –  Elmar Zander Mar 5 '13 at 18:18

Note if $y=1$ then $f(x,1) = x+1$ so let $x = 83$.

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This is a solution, but there might be more. –  Ross Millikan Mar 5 '13 at 17:15
@RossMillikan after your answer, made no sense to rewrite the same thing -- there aren't any more :) –  gt6989b Mar 5 '13 at 19:58

(1,83) and (83,1) seem to be the only answers.

In [1]: for x in range(1,84):
...:     for y in range(1,84):
...:         if x**y + y**x == 84:
...:             print (x,y)
...:
(1, 83)
(83, 1)

(Don't know whether python scripts are allowed to prove things on MSE, though)

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They are a fine approach if the problem is susceptible to exhaustion, as it is here. –  Ross Millikan Mar 5 '13 at 18:10