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 .

Interesting problem!