Knowing $\sum_{k=1}^n k^{2p+1}=\frac{n(n+1)}{2}Q$ (by Faulhaber), can we show $Q$ is odd when $n=4m+1$ or $n=4m+2$? According to Faulhaber's theorem, It is known that the $nth$ triangular number is always a factor of the sum of odd powers of the first $n$ natural numbers, i.e.
If $p>0$, then
$$\sum_{k=1}^n k^{2p+1} = \left(\frac{n(n+1)}{2}\right)Q$$ for some integer $Q$.
My question is:

Is there a way to show that $Q$ is always an odd number if $n=4m+1$ and $n=4m+2$?

I am asking this question because I arrived at this equation
$$\sum_{k=1}^n k^{2p+1} = \left(\frac{n(n+1)}{4}\right)(Q-1)$$ 
Which is easy to see that
$$\left(\frac{n(n+1)}{4}\right)(Q-1)$$ is always an integer when
$n=4m+3$ and $n=4m$ 
Since $Q$ is some integer. 
But $Q$ will have to be an odd number if
$$\left(\frac{n(n+1)}{4}\right)(Q-1)$$
When
$n=4m+1$ and $n=4m+2$.
I have tried all I could to get started but I don't know where to start. Any clue on how to go about it will be appreciated. Thanks.
 A: A proof goes like this:
Observe that $y:=\frac{n(n+1)}{2}$ is odd iff $n\equiv 1  \bmod 4$ or $n\equiv 2 \bmod 4$.
Let $T_p(y):= Q$. We want to show that $T_p(y)$ is odd when $y$ is odd. This is true for $p=1$ since $T_1(y)=y$. We will then use an induction argument:
We have
$$ T_p(y):= \frac{2^py^p}{p+1} -\sum_{i\ge 1}\frac{1}{2i+1}T_{p-i}(y){p\choose 2i} $$
see the proof at  this answer.
Then $$T_p(y)\prod_{i=1}^{\frac{p}{2}}(2i+1)=\frac{2^py^p}{p+1}\prod_{i=1}^{\frac{p}{2}}(2i+1)-\sum_{i\ge 1}T_{p-i}(y){p\choose 2i}\prod_{j=1,j\neq i}^{\frac{p}{2}}(2j+1).$$
we suppose (induction hypothesis) that for $1\le i\le p-1$, $T_{p-i}(y)$ is an odd integer, and we argue that there is an odd number of odd numbers on the right hand side of the above equation and we are done: observe that the first term $\frac{2^py^p}{p+1}\prod_{i=1}^{\frac{p}{2}}(2i+1)$ on the rhs is an even integer as soon as $p\gt 1$ and that there are an odd number of odd numbers in the $\sum$ :because of the symmetry of the binomial coefficients they can be paired except one which is odd : this is ${p\choose p}=1 $ when $p$ is even and when $p$ is odd, this is ${p\choose p-1}=p$.
