Integers to the the n-th power

**leilani22** · October 18th 2010, 06:40 AM

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.

**melese** · October 20th 2010, 08:11 AM

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 $n=<br /> p_1^{n_1}\codt p_2^{n_2}\cdots p_r^{n_r}$ 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!

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Chinese Remainder Theorem.

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