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Thread: Integers to the the n-th power

  1. #1
    Oct 2010

    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.
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  2. #2
    Jun 2010

    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=<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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