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y=3^k+5^k (i)
equals 8 i.e. 2^3 when k=1

Are there additional integer values for k causing y of equation (i) to become a power of an integer?

List those ks or prove there are none.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 Partial solution based on rewording the problem Comment 1 of 1
I'm guessing the problem is asking whether y is a second or higher power.  Otherwise y^1=y so every integer is a power of itself.

By looking at modulos of 3^k and 5^k it is clear that for even values of k, y will have exactly one factor of 2 in its prime factorization so no solutions exist with even k.
If k is odd, y will have exactly three factors of 2.  ie y=2^3*x which means x would have to be a perfect cube of a number with no factors of 2, 3 or 5.  This doesn't seem likely but I'm not sure how to prove it.

If the problem is asking if y can be the power of a prime the above argument shows that it clearly cannot.

 Posted by Jer on 2017-01-23 10:26:10

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