For a positive interger $n$, let $r(n)$ be the sum of the remainders when $n$ is divided by $1,2, \ldots ,n$ respectively. Prove that $r(k)=r(k+1)$ for infinitely many positive intergers $k$.
Here's my try:
By the division algorithm we have $a=bq+r$ and $a-bq=r$. So $r(k)= k(k-1)-(2q_2 + 3q_3+ \ldots kq_k$) and $r(k+1)= (k-1)(k+1) -(2p_2 + 3p_3+ \ldots kp_k$) where $q_i = \lfloor \frac {k}{i} \rfloor$ and $p_i= \lfloor \frac {k+1}{i} \rfloor$
So if $r(k)=r(k+1)$ then $0= k-1 -( 2(p_2-q_2)+ 3(p_3-q_3) \ldots + k(p_k - q_k))$ and I think that $p_n-q_n$ can be equal to $0$ or $1$ depending on if $k+1$ is or is not divisible by $n$. I don't know if my last claim is correct but I think it is.
I would like to receive any hints or help. Thanks in advance.