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An equation (Posted on 2014-07-03) Difficulty: 3 of 5
The equation xxyy=zz has trivial integer solutions if x=1 or y=1. What is the smallest solution to xxyy=zz where x, y, and z are integers and 1<x≤y?

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Some Thoughts Part of it | Comment 2 of 5 |

Let 1<x=y.

Then a^z*a^z=z^z for some a,z, and if so, z=a^2:

a^(a^2)*a^(a^2)=(a^2)^(a^2);

so a^(a^2) = x^x; x is a multiple of a, say k*a

But (a^(a^2))^(1/(k*a)) = k*a yields a^(a/k) = ak, which does not produce non-trivial integer solutions. So x cannot equal y.

Along similar lines, let z^z=a^zb^z, with b exceeding a. Since z=ab,

a^(a^2+an)*(a+n)^(a^2+an) =(a^2+an)^(a^2+an); eligible candidates for z^z soon become large.

a^z and b^z cannot be decomposed so that a^z=x^x and b^z = y^y:

Say b^z=d^d; then for some x, b^x=d since every factor of d is in b. Now (a+n)^(a^2+an) = ((a+n)^x)^((a+n)^x)

Only the powers need be considered:  (a(a+n)) = (x(a+n)^x); but then a = x(a+n)^(x-1), and both a and x must = 1.

So if there is a solution at all, it involves switching factors between a^zb^z to produce x^xy^y; a restricted exercise considering that the latter both have to remain perfect powers after the transfer.


Edited on July 5, 2014, 3:25 am
  Posted by broll on 2014-07-05 03:02:24

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