Possible error in the solution write up for the problem of evaluating $\int_{0}^{1} (-1)^{[nt]}(-1)^{[mt]}dt$ if $m$ and $n$ are positive integers I have a question which I posted here yesterday Question from the write up of a solution concerning integration of greatest integer function.

For $m$ and $n$ positive integers, evaluate $\int_{0}^{1} (-1)^{[nt]}(-1)^{[mt]}dt$ where $[\text{ }]$ denote the greatest integer function.

The first two lines of the solution begin with:

Let $I(n,m)=\int_{0}^{1} (-1)^{[nt]}(-1)^{[mt]}dt$.  We observe that for $0 \leq t \leq \frac{1}{2}$, if $nt$ is not an integer, we have $(-1)^{[nt]}=(-1)^{n+1}(-1)^{[n(1-t)]}$ from which it follows that $I(n,m)=0$ if $n$ and $m$ have opposite parity.

In that post, i should have been more explicit what was it I am confused about.  My apologizes for double posting.  
The issues I am having with are as follows, the sentence: "We observe that for $0 \leq t \leq \frac{1}{2}$, if $nt$ is not an integer, we have $(-1)^{[nt]}=(-1)^{n+1}(-1)^{[n(1-t)]}$ from which it follows that $I(n,m)=0$ if $n$ and $m$ have opposite parity".  I am not sure if this is a two different statement sentence or a one statement sentence, because it is extremely lengthy because I could either read it as: 
We observe that for $0 \leq t \leq \frac{1}{2}$, if $nt$ is not an integer, we have $(-1)^{[nt]}=(-1)^{n+1}(-1)^{[n(1-t)]}$
then
it follows that $I(n,m)=0$ if $n$ and $m$ have opposite parity.
Another thing is, I don't think the identity:  $(-1)^{[nt]}=(-1)^{n+1}(-1)^{[n(1-t)]}$, is correct.  I can interpret it as whether $(-1)^{[a]}$ is either positive or negative depending on whether $[a]$ is respectively even or odd. But ${[nt]}\neq {n+1} + {[n(1-t)]}$. This can be demonstrated for $n=10$ and $t=\frac{1}{3}$, then we have ${[10\frac{1}{3}]}=3\neq {10+1} + {[10(1-\frac{1}{3})]}=11+[\frac{20}{3}]=17$.  Could the author have meant something else instead, like $(-1)^{[\text{negative/positive}]} = (-1)^{[\text{negative/positive}]}(-1)^{[\text{negative/positive}]}$, if so, how did the author came up with the formula.  Also from here since the identity is incorrect, can it still be deduce that "$I(n,m)=0$ if $n$ and $m$ have opposite parity"  
Thank you in advance.
 A: First, we'll prove $(-1)^{[nt]} = (-1)^{n+1}(-1)^{[n(1-t)]}$.
For this, we should take a closer look at $[n(1-t)] = [n - nt]$
If $nt$ isn't an integer, we can see that $[n - nt] = n - [nt] - 1$ (1)
(Further explanation, $n - [nt]$ is the ceiling of $n - nt$ which is a number we know isn't an integer. Thus, $[n - nt] = \lceil(n-nt)\rceil - 1 = n - [nt] - 1$)
Now, using (1)
$(-1)^{n+1}(-1)^{[n(1-t)]} = (-1)^{n+1}(-1)^{n - [nt] - 1} = (-1)^{-[nt]} = (-1)^{nt}$
Now proving $I(m,n) = 0$ if $n$ and $m$ have opposite parity,
Let's simplify the integrand,
$(-1)^{[nt]}(-1)^{[mt]} = (-1)^{n+1}(-1)^{[n(1-t)]}(-1)^{m+1}(-1)^{[m(1-t)]} = (-1)^{m+n+2}(-1)^{[n(1-t)]}(-1)^{[m(1-t)]}$
$m+n$ is odd, so the integrand becomes $I(t) = -(-1)^{[n(1-t)]}(-1)^{[m(1-t)]}$ (2)
If we show that $I(1-t) = -I(t)$ the proof will be completed.
$I(1-t) = -(-1)^{[nt]}(-1)^{[mt]} = -(-1)^{n+1}(-1)^{[n(1-t)]}(-1)^{m+1}(-1)^{[m(1-t)]} = -(-1)^{m+n+2}(-1)^{[n(1-t)]}(-1)^{[m(1-t)]} = -(-1)(-1)^{[n(1-t)]}(-1)^{[m(1-t)]} = (-1)^{[n(1-t)]}(-1)^{[m(1-t)]} = -I(t)$
The proof is complete.
