2015 AIME I Problems/Problem 3
There is a prime number such that is the cube of a positive integer. Find .
Let the positive integer mentioned be , so that . Note that must be odd, because is odd.
Rearrange this expression and factor the left side (this factoring can be done using , or synthetic divison once it is realized that is a root):
Because is odd, is even and is odd. If is odd, must be some multiple of . However, for to be any multiple of other than would mean is not a prime. Therefore, and .
Then our other factor, , is the prime :
Since is odd, let
We know p is a prime number and apparently not an even number. and is an odd number, so a must equal 8.
so we get
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