I'm studying P. Erdős & J. Surányi's book Topics in Theory of Numbers. The Exercise 2.20b) reads:

Prove that for an arbitrary positive integer $N$, there exists $N$ consecutive integers, none of which is a power of an integer (where the exponent is greater than one).

Context: The section in which the exercise is uses the Chinese remainder theorem to prove that for any integer $N>0$ there exists a prime $p$ such that $p-1,\ldots, p-N$ and $p+1,\ldots,p+N$ are all composite. As a matter of fact, exercise 2.20a asks for a proof that there are $N$ consecutive, non squarefree numbers, which I have solved.

Thoughts & Tries: With the idea of using ChRT in mind, I have looked for the appropiate moduli. I have thought on using squares, something like $x_j\equiv j\pmod {p_j^2}$, but the fact that at a number is not the power of $N$ primes does not imply that it is not a power of the next one...

I have had some other ideas, like beginning at $2^{N!}$ or $N^N$, but none of them seems to work.

I still want to solve the problem, so I prefer hints than complete solutions.

  • $\begingroup$ Nice question! Would be easier to solve if you changed "none of which is" to "one of which is not" (though I guess probably less interesting)... $\endgroup$ – barak manos Dec 29 '16 at 17:40
  • $\begingroup$ @barakmanos that's just equivalent to showing at least one non-perfect power exists. $\endgroup$ – Jorge Fernández Hidalgo Dec 29 '16 at 17:42
  • $\begingroup$ @JorgeFernándezHidalgo: So??? $\endgroup$ – barak manos Dec 29 '16 at 17:43
  • $\begingroup$ so it's trivial. $\endgroup$ – Jorge Fernández Hidalgo Dec 29 '16 at 17:45
  • $\begingroup$ @ajotatxe $\Bbb Z[x]$ is not euclidean because fractions do not exist in $\Bbb Z$? How about $\Bbb Z_p[x]$ (padics)? $\endgroup$ – Turbo Jan 5 '17 at 15:36

you can do this easily with the chinese residue theorem.

Just take $n$ prime numbers $p_0,p_1,\dots,p_{n-1}$ and find a number that is a solution to:

$x\equiv 0 \bmod p_0^2$

$x\equiv p_1 -1 \bmod p_1^2$

$x\equiv p_2-2 \bmod p_2^2$


$x\equiv p_{n-1}-(n-1)\bmod p_{n-1}^2$

None of the numbers $x,x+1,\dots,x+n-1$ is a perfect square, as $p_j| x+j$ but $p_j^2 \nmid x+j$.

  • $\begingroup$ How do you show that $x$ is not the power of some other prime (or number)? $\endgroup$ – ajotatxe Dec 29 '16 at 17:30
  • $\begingroup$ added. ${}{}{}{}{}$ $\endgroup$ – Jorge Fernández Hidalgo Dec 29 '16 at 17:31
  • $\begingroup$ If a number $y$ is a perfect $w$'th power then all of the exponents in its prime factorization must be multiples of $w$. $\endgroup$ – Jorge Fernández Hidalgo Dec 29 '16 at 17:32
  • $\begingroup$ Don't you mean $x\equiv p_i-i \mod {p_i^2}$? $\endgroup$ – Darth Geek Dec 29 '16 at 17:33
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    $\begingroup$ Oh yeah, it should be $x\equiv p_j-j\bmod p_j^2$, that way we have $x+j\equiv p_j \bmod p_j^2$ $\endgroup$ – Jorge Fernández Hidalgo Dec 29 '16 at 17:34

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