# Integers to the the n-th power

• October 18th 2010, 06:40 AM
leilani22
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.
• October 20th 2010, 08:11 AM
melese
Chinese Remainder Theorem.
The hint suggests a sufficient form for $n$ that may satisfy the requirements of the problem.

Suppose that $n=2^x\cdot5^y$, where $x$ and $y$ are positive integers.

Now, I will apply the following result: Let $n=
p_1^{n_1}\codt p_2^{n_2}\cdots p_r^{n_r}$
be the prime factorization of $n$. Then $n$ is a $k$-th power if and only if each exponent $n_i$ is divisible by $k$.

$40n=2^{x+3}\cdot5^{y+1}$; $500n=2^{x+2}\cdot5^{y+3}$; and $200n=2^{x+3}\cdot5^{y+2}$.

We construct a system of congruence for $x$.
Since $40n$, $500n$, and $200n$ are a fifth, a sixth, and a seventh powers, respectively, the result implies that $x+3\equiv0\bmod5$, $x+2\equiv0\bmod 6$, and $x+3\equiv0\bmod 7$. Equivalently, we try to solve $x\equiv-3\bmod5$, $x\equiv-2\bmod 6$, and $x\equiv-3\bmod 7$; thus, all solutions are given by $x\equiv172\bmod210$. Check!

Similarly for $y$, construct a system of three congruences and then solve it. I've found that $y\equiv159\bmod210$. Check!

There are, in fact, infinitely many solutions for $n$.
Example: Take $x=172$ and $y=369$ $(369=169+210)$. Here $n=2^{172}\cdot5^{369}$. Indeed, $40n=(2^{35}\cdot5^{74})^5$; $500n=(2^{29}\cdot5^{62})^6$; and $200n=(2^{25}\cdot5^{53})^7$.

The smallest possible $n$ is obtained by taking $x=172$ and $y=159$; here $n=2^{172}\cdot5^{159}$.

Interesting problem!