Nice, but unfortunately, exactly the method I used for solving the problem with integer parts from oral Ecole MInes (that I posted some time ago and after a solution ) works here. Too bad, otherwise it would have been nicer...

Is something wrong in this exercice? If {km/n} and [km/n] are the integral part Take m=n+1 I thing you miss n>m.

For $m=n+1$ it is easy to find that the sum is $\frac{1}{2}$.

AYMANE wrote: Is something wrong in this exercice? If {km/n} and [km/n] are the integral part Take m=n+1 I thing you miss n>m. Nothing is wrong in this exercise. $x = [x]+\{x\}$ for any real number $x$.

m=4 and n=3 And the sum is 1/6-1+2=7/6! Help me

AYMANE wrote: m=4 and n=3 And the sum is 1/6-1+2=7/6! Help me You are not doing the computations correctly.

Thank you i thought that {} means integral part . Sorry

Let $r_a$ be the only integer from $0,1,2,3...,n-1$ such that $n|a-r_a$. We always have: $\left[\frac{mk}{n}\right]=\frac{mk}{n}-\frac{r_{mk}}{n}=r_{mk}$ mod 2, since $m$ is even and $n$ is odd. We also have that $\left\{\frac{mk}{n}\right\}=r_{mk}/n$ and finally we have that: $m,2m,3m,4m.....(n-1)m$ is a complete set of residues modulo $n$. Using this the problem is easy...and only computations are left...

Notice that $mk=qn+a$ where q and a are nonnegative integers ,since m is even and n is odd,so we can know that for even numerator we have + odd numerator we have - Then we can get answer $\frac{1}{2}$

Valentin Vornicu wrote: Let $m,n$ be co-prime integers, such that $m$ is even and $n$ is odd. Prove that the following expression does not depend on the values of $m$ and $n$: \[ \frac 1{2n} + \sum^{n-1}_{k=1} (-1)^{\left[ \frac{mk}n \right]} \left\{ \frac {mk}n \right\} . \] Bogdan Enescu Easy problem, Here's my solution:- I will prove that the fixed value is $\tfrac{1}{2}$.Indeed note $mk\equiv n$ forms a complete residue set(except 0 trivially) on ranging $k$ from $1$ to $n-1$.Hence $$ \sum^{n-1}_{k=1} (-1)^{\left[ \frac{mk}n \right]} \left\{ \frac {mk}n \right\} = \sum^{n-1}_{k=1} (-1)^{\dfrac{mk-mk\mod n}{n}} \left(\dfrac{mk \mod n}{n}\right)= \sum^{n-1}_{k=1} (-1)^k\left(\dfrac{k}{n}\right)=\dfrac{n-1}{2n}$$Thus $\dfrac{n-1}{2n}+\dfrac{1}{2n}=\dfrac{1}{2}$ so we are done.$\blacksquare$ .