equals 8 i.e. 2^3
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.
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