TITLE: Find an integer coprime to sequence sum
QUESTION [9 upvotes]: Given two positive integers $a<b$, can we always find an integer $c\in [a, b]$ that is coprime to $\sum_{a\le i\le b} i$?

REPLY [7 votes]: I believe a large counterexample exists
Actually, no, there is no counterexample. Let's consider several cases:
Case 1: $b-a$ is even. Then the progression has odd number $2X+1$ of terms from $A-X$ to $A+X$ and the sum is $A(2X+1)$. Now let $p_1<p_2<\dots<p_n$ be all (automatically odd) primes that divide $2X+1$ but not $A$ (at least one such prime exists because otherwise we can just take $c=A+1$). Consider the numbers $A\pm p_1\dots p_{n-1}$. Note that they are in the range $[A-X,A+X]$ and the only common prime factor they may have with $A(2X+1)$ is $p_n$. But they cannot be divisible by $p_n$ simultaneously, so one of them can be taken as $c$.
Case 2: $b-a$ is odd, $b=A+X,a=A-X+1$, $X$ is odd. Then the sum is $(2A+1)X$. Again, let $p_1<p_2<\dots<p_n$ be all (automatically odd) primes that divide $X$ but not $2A+1$. This time just consider the number 
$\frac{(2A+1)+ p_1\dots p_{n}}{2}$.
Case 3: $b-a$ is odd, $b=A+X,a=A-X+1$, $X$ is even. Then the sum is still $(2A+1)X$. Again, let $p_1<p_2<\dots<p_n$ be all odd primes that divide $X$ but not $2A+1$. Consider the numbers $\frac{(2A+1)\pm p_1\dots p_{n}}{2}$. One of them is now even and the other one is odd. Also they are both in the range $[A-X+1,A+X]$. So, the odd one can be taken as $c$.