Prove that $\mu(n)\mu(n+1)\mu(n+2)\mu(n+3)=0$ for $n\ge 1$

I got this question that asks

Prove that $$\mu(n)\mu(n+1)\mu(n+2)\mu(n+3)=0$$ where $\mu$ is the Mobius function.

So, basically, we need to prove that out of every four consecutive integers, atleast one is not square-free.

I tried using Induction. So, for $n=1$, the given equality is clearly true. If we assume that it's true for some $n=k$, then either one (or more) of $\mu(k)$, $\mu(k+1)$, $\mu(k+2)$ or $\mu(k+3)$ is $0$. If it's the last three, the $n=k+1$ case immediately follows. If not, then we need to show that $\mu(k+4)=0$. This is where I am stuck.

Solution 1:

Hint: Among every $n$ consecutive integers, there exists one that is divisible by...