Let $a_1,\cdots, a_k$ be positive real numbers of which at least one is not an integer. Prove that there are infinitely many positive integers $n$ so that $n$ and $\lfloor a_1 n \rfloor +\cdots + \lfloor a_k n\rfloor$ are coprime.
Some basic properties of gcds could be useful: $\gcd(a+kb,b) = \gcd(a,b)$ for an integers $a,k,b$, if $a | bc,$ then $a/d | c,$ where $d = \gcd(a,b).$ If $\gcd(a,c)=\gcd(b,c)=1$, then $\gcd(ab,c)=1$.
There's also the fact that (though this is most likely not useful for the given problem) $\lfloor x\rfloor + \cdots + \lfloor x+(n-1)/n\rfloor = \lfloor nx\rfloor$ for any real number x and positive integer $n$. Indeed if $\dfrac{i}n \leq x-\lfloor x\rfloor < \dfrac{i+1}n,$ then $\lfloor x+\dfrac{k}n\rfloor =\lfloor x\rfloor$ for $0\leq k\leq n-i-1$ (since then $\lfloor x\rfloor + (i+k)/n \leq x+\dfrac{k}n < \lfloor x\rfloor + \dfrac{i+k+1}n$ and for $n-i\leq k\leq n-1,$ it equals $\lfloor x\rfloor$. But $n\lfloor x\rfloor + i\leq nx< n\lfloor x\rfloor + i+ 1,$ so $\lfloor nx\rfloor = n\lfloor x\rfloor + i,$ from which the result follows.
If $a_1, \cdots, a_k$ are all integers, then clearly the two given numbers are not coprime for any $n>1.$ I think a proof by contradiction could be useful. Suppose WLOG that $a_1$ is not an integer. First consider the case where $a_1 = \dfrac{p}q, \gcd(p,q)=1, q > 1,p > 0.$ Then $ a_1 n$ is an integer if and only if n is a multiple of $q$. Otherwise, $\lfloor a_1 n\rfloor = a_1 n – \dfrac{p n\mod q}{q}.$
I'm not sure if it's true that for a sequence of prime numbers $p_n$ going to infinity, $\dfrac{\lfloor a_1 n \rfloor +\cdots + \lfloor a_k n\rfloor }{p_n}$ is a rational sequence converging to $a_1+\cdots + a_k$. I know that if $(a_n)$ is a convergent sequence of real numbers, then $(\dfrac{a_1+\cdots + a_n}n)$ converges to the same value as $(a_n)$.
Best Answer
I think this strategy can produce infinitely many $n$ as desired.
First, notice that since at least one of $a_1, ... , a_k$ is not an integer, we can find infinitely many primes, namely $p$, such that:
$$(a_1+\ ...\ +a_k)p-k\lt [a_1p]+\ ...+[a_kp]\lt (a_1+\ ...\ +a_k)p.$$
Now, let's define $\alpha=a_1+\ ...+ \ a_k.$ There also exist infinitely many prime numbers such that there is an integer between $\alpha -1$ and $\alpha -\frac {k}{p}$. In other words,
$$\alpha -1\le m \lt \alpha -\frac {k}{p}, $$
where $m$ is an integer.
Now, assume $n=p$ is a prime number that satisfies both of the conditions mentioned above (there are infinitely many of those prime numbers) then we conclude: $$pm\lt p\alpha-k=(a_1+\ ...\ +a_k)p-k\lt [a_1p]+\ ...+[a_kp]\lt (a_1+\ ...\ +a_k)p=\alpha p\le pm+p.$$
Hence $[a_1p]+\ ...+[a_kp]=pm+s$, where $1\le s \le p-1$. This shows that $n=p$ and $[a_1p]+\ ...+[a_kp]$ are coprime.