TITLE: Prove there are k consecutive non-squarefree integers
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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.

REPLY [3 votes]: 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.