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I need to show that there exist three consecutive numbers, where each of them is divisible by the 1999th power of an integer. So far, I've taken $a_1 = x-1, a_2 = x$ and $a_3 = x + 1$, and derived a system of congruences: $$\begin{align}x \equiv -1\bmod{n^{1999}} \\ x \equiv 0\bmod{n^{1999}} \\ x\equiv 1\bmod{n^{1999}}\end{align}$$

But I am unsure of how to proceed. A hint would be greatly appreciated.

Edit: As mentioned in the comments and answer, the said powers can be of any three distinct integers.

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    Your wording is not clear. There exist integers $n_1,n_2,n_3$ so that $n_1^{1999}\mid a_1$, $n_2\mid a_2^{1999}$, and $n_3^{1999}\mid a_3$. [Unless you are allowed to take $n=1$, an easy solution.] – GEdgar Aug 13 '22 at 00:39
  • i considered that, but the question unfortunately isn't too clear about this because it just says "integers". so is it fair to assume that it means integers excluding 1 and -1? @GEdgar –  Aug 13 '22 at 00:46
  • Presumably, under $1999$th powers, you don't allow $1.$ – Thomas Andrews Aug 13 '22 at 00:49
  • It's not possible for three consecutive numbers to have the same factor, other than $1.$ – Thomas Andrews Aug 13 '22 at 00:50
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    Your congruence has each of them divisible by the same $1999$th power, but the text version at the beginning of the question does not say the same $1999$th power. Your title rephrases the question as if $n$ is the same, too, for each $i.$ – Thomas Andrews Aug 13 '22 at 00:51

2 Answers2

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Just show that there is an $x$ such that: $$x\equiv 1\pmod {2^{1999}}\\x\equiv 0\pmod {3^{1999}}\\x\equiv-1\pmod{5^{1999}}.\tag1$$

You have assumed the $n$ is the same, but that is nowhere stated in the text of the question, and it is not possible to solve it that way except with $n =\pm 1,$ which trivially works, but is probably implicitly excluded.

You can, of course, use any three relatively prime $1999$th powers in $(1).$

Thomas Andrews
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Since $\Bbb Z_{p^{1999}}\otimes\Bbb Z_{q^{1999}}\otimes\Bbb Z_{r^{1999}}$ is canonically isomorphic to $\Bbb Z_{(pqr)^{1999}}$, by the Chinese remainder theorem, we are guaranteed a unique solution $\pmod{(pqr)^{1999}}$, corresponding to $(1,0,-1)\in\Bbb Z_{p^{1999}}\otimes\Bbb Z_{q^{1999}}\otimes\Bbb Z_{r^{1999}}$.

calc ll
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