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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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solution | Comment 3 of 5 |
If x and y have no common factor >1 then both factors on LHS are z-th powers which means x and y are multiples of z, say x=az, y=bz.

Substitute those values and take the z-th root to get (az)^a + (bz)^b = z which requires a=b=1, x=y=z=1.  This is the solution given.

Otherwise, gcf(x,y)=d>1.

Let x=dp, y=dq, z=dr, gcf(p,q)=1.

Substitute and take the d-th root to get d^(p+q) * p^p * q^q = d^r * r^r.

The powers of d must be equal so factor them out leaving p^p * q^q = r^r which reverts to the first case.

So (x,y,z)=(1,1,1) is the only solution.


  Posted by xdog on 2014-07-13 10:40:04
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