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So, I've got a question for class that asks me to prove the existence of arbitrarily long runs of consecutive integers where $\mu(n)$ is zero.

I've started the proof, but I need a bit of help midway through.

Assume there exists a run of length n, which we currently assume is the longest possible chain.

If we induct on n, we can assume there is a run of the form $m_1p_1^2, m_2p_2^2, ..., m_np_n^2$ (where each of the $p_i$ are prime).

If I add $M = lcm(m_1p_1^2, m_2p_2^2, ..., m_np_n^2)$ to each number, I get another run of length n.

So here's where the issue starts. When I talked to my advisor about it, he referenced a theorem where there exists a prime $p$ congruent to $1\ mod\ M$.

I am unfamiliar with this theorem, so I'm not sure how to use this information. Any chance I can get some help? Or else is there a simpler way to go about this problem?

Barney
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  • This is exercise 2.3.18 of Niven, Zuckerman & Montgomery's An Introduction to the Theory of Numbers. – 19021605 Sep 19 '25 at 06:15

3 Answers3

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Simpler solution: we seek $n$ to satisfy all of the following:

$$n\equiv 0\pmod{p_1^2}$$

$$n+1\equiv 0\pmod{p_2^2}$$

$$n+2\equiv 0\pmod{p_3^2}$$

$$\vdots$$

$$n+k\equiv 0\pmod{p_{k+1}^2}$$

Now we use the Chinese remainder theorem.

tox123
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vadim123
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  • Thanks for the help! I really appreciate it. – Barney Nov 20 '14 at 20:47
  • @vadim123 Could you possibly explain this in more detail? I’m not understanding the argument, I guess. – BalsamicVinegar Feb 12 '20 at 23:56
  • The CRT will give a value $n$ that simultaneously satisfies all of the modular equations, mod $p_1^2p_2^2\cdots p_k^2$. Hence, $p_1^2$ will divide $n$, so $n$ will not be squarefree. $p_2^2$ will divide $n+1$, so $n+1$ will not be squarefree, etc. – vadim123 Feb 13 '20 at 15:56
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There are two relevant theorems at play. A theorem that guarantees a prime congruent to $1 \bmod M$ is Dirichlet's Theorem on Primes in Arithmetic Progressions, which says that as long as $x$ and $y$ are coprime, then the sequence $x + ny$ as $n$ increases contains infinitely many primes.

The second theorem at play is the Chinese Remainder Theorem, which allows you to skip the induction and directly prove the result.

davidlowryduda
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  • Thank you! I believe that was just what I needed. I'm going to give the CRT version a chance as well, since that one definitely seems much more intuitive. – Barney Nov 20 '14 at 20:46
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Instead of induction, how about using the Chinese Remainder Theorem to find a number that is $n$ modulo $(p_n)^2$ for $1\le n\le k$?

hmakholm left over Monica
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