Consecutive Numbers pt. 3

Number Theory Level 2

For every positive integer $n,$ does there always exist $n$ consecutive positive integers such that none of them can be written in the form $m^k,$ where $m$ and $k$ are positive integers with $k > 1$ ?

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1 solution

Steven Yuan
Jul 20, 2017

Let $p_1, p_2, \dots, p_n$ be distinct prime numbers. To prove that there exists $n$ consecutive integers that are not perfect powers, we can show that the $i$ th number is congruent to $p_i$ modulo $p_i^2,$ since that would imply that the exponent on $p_i$ is exactly 1, which disqualifies the number from being a perfect power.

Thus, we must find a positive integer $x$ such that $x + i \equiv p_i \! \pmod{p_i^2}$ for $i = 1, 2, \dots, n.$ By the Chinese Remainder Theorem, there exists a solution $x$ to this system. This choice of $x$ guarantees that all of $x + 1, x + 2, \dots, x + n$ are not the perfect power of any positive integer. $\blacksquare$

Consecutive Numbers pt. 3

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