Given positive integers $a,b,c$ which are pairwise coprime. Let $f(n)$ denotes the number of the non-negative integer solution $(x,y,z)$ to the equation $$ax+by+cz=n.$$ we have Prove that there exists constants $\alpha, \beta, \gamma \in \mathbb{R}$ such that for any non-negative integer $n$, $$|f(x)- \left( \alpha n^2+ \beta n + \gamma \right) | < \frac{1}{12} \left( a+b+c \right).$$
proof: Let $\zeta_a=e^{2\pi i/a}$ etc., then all the $\zeta_a^j\,(1\leq j\leq a-1)$ and $\zeta_b^j\,(1\leq j\leq b-1)$ and $\zeta_c^j\,(1\le j\le c-1)$ are different, so we can calculate $$\frac{1}{(1-x^a)(1-x^b)(1-x^c)}=\frac{\lambda_2}{(1-x)^3}+\frac{\lambda_1}{(1-x)^2}+\frac{\lambda_0}{1-x}+\sum_{j=1}^{a-1}\frac{\rho_j(a)}{1-\zeta_a^jx}+\sum_{j=1}^{b-1}\frac{\rho_j(b)}{1-\zeta_b^jx}+\sum_{j=1}^{c-1}\frac{\rho_j(c)}{1-\zeta_c^jx},$$ where $\lambda_i\,(0\leq i\leq 2)$ are constants depending only on $(a,b,c)$, and $$\rho_j(a)=\lim_{x\to\zeta_a^{-j}}\frac{1-\zeta_a^jx}{(1-x^a)(1-x^b)(1-x^c)}=\frac{1}{a}\frac{1}{(1-\zeta_a^{-jb})(1-\zeta_a^{-jc})},$$ and similar for $\rho_j(b)$ and $\rho_j(c)$. Now the $n$-th power coefficient $$[x^n]\bigg(\frac{\lambda_2}{(1-x)^3}+\frac{\lambda_1}{(1-x)^2}+\frac{\lambda_0}{1-x}\bigg)=\lambda_2\binom{n+2}{2}+\lambda_1\binom{n+1}{1}+\lambda_0=\alpha n^2+\beta n+\gamma,$$ so we just need to estimate the $n$-th power coefficient $$[x^n]\bigg(\sum_{j=1}^{a-1}\frac{\rho_j(a)}{1-\zeta_a^jx}+\sum_{j=1}^{b-1}\frac{\rho_j(b)}{1-\zeta_b^jx}+\sum_{j=1}^{c-1}\frac{\rho_j(c)}{1-\zeta_c^jx}\bigg):=\Delta_n$$ and prove $|\Delta_n|<\frac{1}{12}(a+b+c)$. Clearly $$\Delta _n=\frac{1}{a}\sum_{j=1}^{a-1}\frac{\zeta_a^{jn}}{(1-\zeta_a^{-jb})(1-\zeta_a^{-jc})}+(\mathrm{two\ other\ similar\ terms)},$$ so it suffices to show that $$\bigg|\sum_{j=1}^{a-1}\frac{\zeta_a^{jn}}{(1-\zeta_a^{-jb})(1-\zeta_a^{-jc})}\bigg|\leq\frac{a^2}{12}.$$ By Cauchy-Schwartz the above is bounded by $$\bigg(\sum_{j=1}^{a-1}|1-\zeta_a^{-jb}|^{-2}\bigg)^{1/2}\bigg(\sum_{j=1}^{a-1}|1-\zeta_a^{-jc}|^{-2}\bigg)^{1/2}=\frac{1}{4}\bigg(\sum_{j=1}^{a-1}\frac{1}{\sin^2(\pi jb/a)}\bigg)^{1/2}\bigg(\sum_{j=1}^{a-1}\frac{1}{\sin^2(\pi jc/a)}\bigg)^{1/2},$$ moreover when $1\leq j\leq a-1$ varies, $\{jb\}$ and $\{jc\}$ each runs over a complete residue class modulo $a$ (except $0$), so it suffices to show that $$\sum_{j=1}^{a-1}\frac{1}{\sin^2(\pi j/a)}<\frac{a^2}{3}.$$ However by considering the roots of the polynomial $$P_a(\sin x)=\frac{\sin(ax)}{\sin x}\mathrm{\ for\ }a\mathrm{\ odd},\quad P_a(\sin x)=\frac{\sin(ax)}{\sin x\cos x}\mathrm{\ for\ }a\mathrm{\ even}$$ and using Vieta's theorem, we can easily calculate $$\sum_{j=1}^{a-1}\frac{1}{\sin^2(\pi j/a)}=\frac{a^2-1}{3}.$$ The proof is now complete.
**My conjecture ** : Given positive integers $a,b,c,d$ which are pairwise coprime. Let $f(n)$ denotes the number of the non-negative integer solution $(x,y,z,w)$ to the equation $$ax+by+cz+dw=n.$$ we have Prove that there exists constants $\alpha, \beta, \gamma ,\delta\in \mathbb{R}$ such that for any non-negative integer $n$, $$|f(x)- \left( \alpha n^3+ \beta n^2 + \gamma n+\delta\right) | < A\left( a+b+c+d \right).$$
and I don't this result is odd or new,if you know,can you tell me,Thanks
