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Thread: Integers to the the n-th power

  1. #1
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    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
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    Chinese Remainder Theorem.

    The hint suggests a sufficient form for $\displaystyle n $ that may satisfy the requirements of the problem.

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

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

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

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

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

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

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

    Interesting problem!
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