Conjecture about $f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(P_r(n) \frac{a \pi}{b}\right) $ I asked another question related this question. $r=1$ was considered in the related question.You may see proofs for $r=1$.
I would like to generalize the conjecture when $r$ is any positive integer in this question.
Generalized Conjecture:
$$f(m)=\sum\limits_{n = 1 }^ m (-1)^n  \sin\left(P_r(n) \frac{a \pi}{b}\right) \tag 1 $$
I have a conjecture that if $P_r(n)=\sum\limits_{k = 1 }^ n k^{2r}$ where r is a positive integer,
$f(m)$ function is periodic function when $a,b,m$ positive integers and
$ \sum\limits_{k = 1 }^T f(k)=0 $
where ($T$) is the period value. 

I tested a lot of polynomials that is different than $P_r(n)$ but they failed in my tests.
I have not found any polynomial which is different from $c.P_r(n)$ that satisfy   $\sum\limits_{k = 1 }^T f(k)=0 $ for all $a,b,m$ positive integers and c is a rational number.  


*

*Please help me how the generalized conjecture can proven or disproved . 

*Please find a counter_example polynomial that is different from $c.P_r(n)$ that satisfies $\sum\limits_{k = 1 }^T f(k)=0 $ for all possible $a,b,m$ positive integers and c is a rational number.  .


Thanks a lot for answers.
Test WolframAlpha link for $P_1(n)=\sum\limits_{k = 1 }^ n k^{2}$
Test WolframAlpha link for $P_2(n)=\sum\limits_{k = 1 }^ n k^{4}$
Test WolframAlpha link for $P_3(n)=\sum\limits_{k = 1 }^ n k^{6}$
Please note that while checking the links, see partial sum graphics in web page for finding period and symmetry while testing some $a,b,m$ values.

My conjecture can be rewritten in the other form as @Gerry Myerson pointed in comment: 
$$f(m)=\sum\limits_{n = 1 }^ m (-1)^n  \sin\left(P_r(n) \frac{a \pi}{b}\right) \tag 2 $$
\begin{align*}
u(m) =  \sum_{n=1}^{m} (-1)^n e^{i P_r(n) \frac{a \pi}{b}}
\end{align*}
\begin{align*}
f(m) = \operatorname{Im}\left( u(m) \right)
\end{align*}
if $P_r(n)=\sum\limits_{k = 1 }^ n k^{2r}$ where r is a positive integer,
$u(m)$ function is periodic complex function when $a,b,m$ positive integers and
$ \operatorname{Im}\left(\sum\limits_{k = 1 }^T u(k)\right)=0 $
where ($T$) is the period value for all $a,b,m$ positive integers. 
EDIT:
I have found out a counter-example and It shows  my generalized conjecture can be extended more. I have tested with many numerical values that It supports my extended conjecture below.
$$f(m)=\sum\limits_{n = 1 }^ m (-1)^n  \sin\left(G(n) \frac{a \pi}{b}\right) \tag 2 $$
$$G(n)=\frac{n(n+1)(2n+1)(3n^2+3n-4)}{30}$$  It also satisfies my generalized conjecture $(1)$ above. $G(n)$ can be written as:
$$G(n)=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}-\frac{3}{5}\frac{n(n+1)(2n+1)}{6}=\sum\limits_{k = 1 }^ n k^{4}-\frac{3}{5}\sum\limits_{k = 1 }^ n k^{2}=P_2(n)-\frac{3}{5}P_1(n)$$
Test link for $a=2,b=5$ and $m=500$
The numerical values and my works on the subject estimates the extension of the conjecture above. It is just strong sense without proof that it must be true.
More generalized conjecture can be written:
Extended Conjecture:
$$f(m)=\sum\limits_{n = 1 }^ m (-1)^n  \sin\left(\sum\limits_{k = 1 }^\infty  \frac{a_k \pi}{b_k}P_k(n) \right) \tag 3 $$
More extended conjecture claims that if $P_r(n)=\sum\limits_{k = 1 }^ n k^{2r}$ where r is a positive integer,
$f(m)$ function is periodic function when $a_k$ is any integers, $b_k$ is non-zero integers and $m$ positive integers.$\sum\limits_{k = 1 }^T f(k)=0 $
where ($T$) is the period value for all possible $a_k,b_k,m$ integers.  

I have been still looking for $G(n)$ polynomials that is different from  $G(n)=\sum\limits_{k = 1 }^\infty  \frac{a_k}{b_k}P_k(n)$ satisfies $ \sum\limits_{k = 1 }^T f(k)=0 $  $\tag{4}$  for all $a,b,m$ positive integers 
(Please consider Equation $(2)$)
Please note that another question has been posted for extended conjecture (Equation ($3$)).
Thanks for answers
 A: $$\mathbf{\color{brown}{Sufficient\ conditions.}}$$

Let $g(x)$ for integer $k,m$ have the properties
  $$g(k+2m)=g(k),\tag1$$ $$g(k)=-g(-k-1),\tag2$$ then
  $$\boxed{\sum\limits_{n=1}^m(-1)^ng(n)=0.}\tag3$$

Really, $(1)-(2)$ leads to 
\begin{align}
&g(n)=-g(-n-1)=-g(2m-n-1),\\
&\sum\limits_{n=1}^m(-1)^ng(n) = -\sum\limits_{n=1}^{m}(-1)^{n}g(2m-n-1) = -\sum\limits_{n=1}^{m}(-1)^ng(n),\hspace{40pt}\\
&\mathbf{\sum\limits_{n=1}^m(-1)^ng(n) = 0.}
\end{align}

$$\mathbf{\color{brown}{The\ periodic\ property.}}$$
Easy to see that the periodic property $(1)$ is satisfied for any function in the form of 
\begin{cases}
g(x)=\sin\frac\pi mxP(x),\\[4pt]
P(x)=\sum\limits_{d=0}^Dp_dx^d,\\[4pt]
p_d\in\mathbb Z.\tag4
\end{cases}
Then, using the binomial formula,
\begin{align}
&g(n+2m) = \sin\left(\frac\pi m\sum\limits_{d=0}^Dp_d(n+2m)^d\right) = \\
&\sin\left(\frac\pi m\sum\limits_{d=0}^Dp_d\left(n^d + 2m\sum\limits_{j=0}^{d}\binom{d}{j+1}n^{d-j-1}(2m)^j\right)\right) = \\[4pt]
&\sin\left(\frac\pi m\sum\limits_{d=0}^Dp_dn^d + 2\pi \sum\limits_{d=0}^Dp_d\sum\limits_{j=0}^{d}\binom{d}{j+1}n^{d-j-1}(2m)^j\right) = \sin\left(\frac\pi m\sum\limits_{d=0}^Dp_dn^d\right),\\[4pt]
&\mathbf{g(n+2m)=g(n).}
\end{align}
$\mathbf{\color{green}{Affect\ of\ the\ multipliers.}}$
If $\gcd\limits_{n=1\dots m} P(n) = 1,$ then the period $T$ of g(n) equals $2m.$
If $\gcd\limits_{n=1\dots m} P(n) = 2p+1 > 1,$ then
\begin{align}
&T=\frac{2m}{2p+1},\\[4pt]
&\sum\limits_{n=1}^m(-1)^ng(n) = \sum\limits_{h=0}^{2p} \sum\limits_{n=1}^T(-1)^{hT+n}g(hT+n) = \sum\limits_{h=0}^{2p}(-1)^{hT} \sum\limits_{n=1}^T(-1)^{n}g(n)\\[4pt]
& = (1+p((T+1)\bmod2))\sum\limits_{n=1}^T(-1)^{n}g(n)\tag5\\[4pt]
\end{align}
$$\mathbf{\color{brown}{Modified\ sufficient\ conditions}}\ (1)-(3)$$
Let $Q(x)=P\left(x+\frac12\right),$ then, using $(2),$
$$Q(-x) = P\left(-x-\frac12\right) = -P\left(x+\frac12-1\right) = -P\left(x-\frac12\right)=Q(-x).\tag6$$
Taking in account $(5)-(6),$ one can rewrite the conditions $(1)(3))$ in the next form.

If $P(n)$ is the odd polynomial with the integer coefficients,
  and $T$ is the minimal period of the sequence 
  $$g_n=\sin\left(\frac\pi mP\left(n+\frac12\right)\right),\tag7$$
  then
  $$\boxed{\sum\limits_{n=1}^T(-1)^{n}g_n=0.}\tag8$$


$$\mathbf{\color{green}{Partial\ solutions.}}$$
For the odd $d,$ one can obtain
\begin{align}
&\mathbf{d=1:}\qquad
\boxed{P_1\left(\frac{n+1}2\right)\sim 2n+1}.\tag9\\
&\mathbf{d=3:}\qquad
\boxed{P_3\left(\frac{n+1}2\right)\sim 2n^3+3n^2+n+c(2n+1)}\dots.\tag{10}\\
\end{align}
Resolving polynomial $(9)\ \mathbf{\color{brown}{\ is\ the\ counterexample.}}$
If $c=0$ then polynomial $(10)$ equals to $6P_1(n)$ from OP .
Easy to see that $\mathbf{\color{brown}{resolving\ polynomials\ are\ additive}}.$ Also, this fact follows from $(7).$
On the other hand, the polynomials $P_r(n)$ for the even $r$ have the required form. So all of them are the solutions too, and the other solutions are the linear combination of known ones. Although, instead of these polynomials can be used monoms in the form of $(2n+1)^{2r+1}$ or linear combinations of them.
