Summation for $m$ and $n$

From the first few $n$ , we note that

$\begin{aligned} S(n) & = (-1)^{n+1}\left \lceil \dfrac n2 \right \rceil = \begin{cases} \dfrac {n+1}2 & \text{if }n \text{ is odd.} \\ - \dfrac n2 & \text{if }n \text{ is even.} \end{cases} \end{aligned}$ .

And we can prove the claim is true for all $n \ge 1$ by induction (see Proof below).

Let us consider the three cases when $m$ and $n$ are both odd, both even and one odd, one even.

Case: both $m$ and $n$ are odd

$\begin{aligned} \mathscr S & = S(m) + S(n) + S({\color{#3D99F6}m+n}) & \small \color{#3D99F6} \text{Note that }m+n \text{ is even.} \\ & = \frac {m+1}2 + \frac {n+1}2 - \frac {m+n}2 \\ & = \color{#D61F06}1 \ne 2011 \end{aligned}$

Case: both $m$ and $n$ are even

$\begin{aligned} \mathscr S & = S(m) + S(n) + S({\color{#3D99F6}m+n}) & \small \color{#3D99F6} \text{Note that }m+n \text{ is even.} \\ & = - \frac m2 - \frac n2 - \frac {m+n}2 \\ & = \color{#D61F06}-(m+n) \ne 2011 \end{aligned}$

Case: $m$ is odd, $n$ one even

$\begin{aligned} \mathscr S & = S(m) + S(n) + S({\color{#3D99F6}m+n}) & \small \color{#3D99F6} \text{Note that }m+n \text{ is odd.} \\ & = \frac {m+1}2 - \frac n2 + \frac {m+n+1}2 \\ & = m + 1 \end{aligned}$

If $\mathscr S = 2011$ , then $m + 1 = 2011$ or $m = 2010$ , which contradicts the assumption that $m$ is odd. Therefore there is no solution.

Therefore, the total numbers of pairs of $(m,n)$ satisfying the equation is $\boxed{0}$ .

$$

---

Proof:

• For $n$ is odd: For $n=1$ , $S_1 = \displaystyle \sum_{k=1}^1 (-1)^{k+1} k = 1 = \frac {1+1}2$ , therefore, the claim is true for $n=1$ .
• Assuming the claim is true for $n_{odd}$ , then

$\begin{aligned} \qquad S({\color{#3D99F6}n+2}) & = \sum_{k=1}^{\color{#3D99F6}n+2} (-1)^{k+1} k \\ & = S(n) - (n+1) + (n+2) \\ & = \frac {n+1}2 - n-1 + n+2 \\ & = \frac {{\color{#3D99F6}(n+2)}+1}2 \end{aligned}$

$\quad$ Therefore the claim is true for $n+2$ and is true for all odd $n$ .

• For $n$ is even: For $n=2$ , $S_2 = \displaystyle \sum_{k=1}^2 (-1)^{k+1} k = 1 - 2 = - 1 = - \frac 22$ , therefore, the claim is true for $n=1$ .
• Assuming the claim is true for $n_{even}$ , then

$\begin{aligned} \qquad S({\color{#3D99F6}n+2}) & = \sum_{k=1}^{\color{#3D99F6}n+2} (-1)^{k+1} k \\ & = S(n) + (n+1) - (n+2) \\ & = - \frac n2 + n +1 - n-2 \\ & = - \frac {\color{#3D99F6}(n+2)}2 \end{aligned}$

$\quad$ Therefore the claim is true for $n+2$ and is true for all even $n$ .

• Therefore, the claim is true for all $n \ge 1$ .

$\large \displaystyle S(n) = 1 - 2 + 3 - 4 + ... + (-1)^n \cdot n$

$\large \displaystyle S(1) = 1, S(2) = -1, S(3) = 2, S(4) = -2...$

$\color{#20A900} \boxed {\large \displaystyle S(n) = (-1)^{n+1} \cdot \left \lceil \frac{n}{2} \right \rceil }$

If both $m$ and $n$ are even, $m+n$ will be also even. Then:

$\large \displaystyle S(m) + S(n) + S(m+n) = -\frac{n}{2} - \frac{m}{2} - \frac{n+m}{2}$

$\color{#20A900} \boxed {\large \displaystyle S(m) + S(n) + S(m+n) = -(m+n)}$

Which will never be equal to 2011, since both $m$ and $n$ have to be positive

If both $m$ and $n$ are odd, $m+n$ will be even. Then:

$\large \displaystyle S(m) + S(n) + S(m+n) = \frac{n+1}{2} + \frac{m+1}{2} - \frac{m+n}{2}$

$\color{#20A900} \boxed {\large \displaystyle S(m) + S(n) + S(m+n) = 1}$

Which obviously will never be equal to 2011.

If one of them, say $n$ , is even, and the other, $m$ is odd, $m+n$ will be odd. Then:

$\large \displaystyle S(m) + S(n) + S(m+n) = \frac{m+1}{2} - \frac{n}{2} + \frac{m+n+1}{2}$

$\color{#20A900} \boxed {\large \displaystyle S(m) + S(n) + S(m+n) = m+1}$

Which will be equal to $2011$ for $m$ (the one set as odd) equal to $\color{#20A900} \boxed {2010}$ , while $n$ (the one set as even) can take any value. But this is a contradiction, since $m$ has to be odd. So, the answer is $\color{#3D99F6} \boxed {\large \displaystyle 0}$ pairs.