Integers to the the n-th power
Find a positive integer n so that 40n is a fifth power (of an integer), 500n is a sixth power, and 200n is a seventh power, or explain why it is impossible to do so. (Hint: Let n=2^x * 5^y. Use the assumptions about n to find systems of linear congruences that x and y must satisfy). You may leave n in terms of its canonical prime factorization.
Chinese Remainder Theorem.
The hint suggests a sufficient form for that may satisfy the requirements of the problem.
Suppose that , where and are positive integers.
Now, I will apply the following result: Let be the prime factorization of . Then is a -th power if and only if each exponent is divisible by .
; ; and .
We construct a system of congruence for .
Since , , and are a fifth, a sixth, and a seventh powers, respectively, the result implies that , , and . Equivalently, we try to solve , , and ; thus, all solutions are given by . Check!
Similarly for , construct a system of three congruences and then solve it. I've found that . Check!
There are, in fact, infinitely many solutions for .
Example: Take and . Here . Indeed, ; ; and .
The smallest possible is obtained by taking and ; here .