2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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:

$x\equiv 0 \bmod p_0^2$

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

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

$\dots$

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


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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.