$0 \leq i \leq \alpha+1$, then $\alpha=4i-1$.

2.  Let $R_{2}=(x^{3}-9xy^{2})(x^{2}-y^{2})^{\alpha}$. If the coefficients of $x^{2\alpha+3-2i}y^{2i}$ in $R_{2}$ are not equal to $0$.

3.  Let $R_{3}=(x^{2}+3y^{2})^{2}(x^{2}-y^{2})^{\alpha}$. If the coefficients of $x^{2\alpha+4-2i}y^{2i}$ in $R_{3}$ are equal to $0$ for $0 \leq i \leq \alpha+2$, then $48\alpha+112$ is a square number.

\(1) We have $$\begin{aligned}
R_{1}&=(x^{2}+3y^{2})(x^{2}-y^{2})^{\alpha} \\
 &=\sum_{i=0}^{\alpha+1} (-1)^{i} \left( \binom{\alpha}{i}-3\binom{\alpha}{i-1} \right)x^{2\alpha+2-2i}y^{2i}.\end{aligned}$$ If the coefficients of $x^{2\alpha+2-2i}y^{2i}$ in $R_{1}$ are equal to $0$, $$\binom{\alpha}{i}-3\binom{\alpha}{i-1}=0.$$ Then we have $$\begin{aligned}
&\frac{\alpha !}{i !(\alpha-i)!}-3\frac{\alpha !}{(i-1) !(\alpha-i+1)!}=0 \\
\Leftrightarrow\ &\alpha-i+1-3i=0 \\
\Leftrightarrow\ &\alpha=4i-1.\end{aligned}$$

\(2) We have $$\begin{aligned}
R_{2}&=(x^{3}-9xy^{2})(x^{2}-y^{2})^{\alpha} \\
 &=\sum_{i=0}^{\alpha+1} (-1)^{i} \left( \binom{\alpha}{i}+9\binom{\alpha}{i-1} \right)x^{2\alpha+3-2i}y^{2i}.\end{aligned}$$ If the coefficients of $x^{2\alpha+3-2i}y^{2i}$ in $R_{2}$ are equal to $0$, $$\binom{\alpha}{i}+9\binom{\alpha}{i-1}=0.$$ Then we have $$\begin{aligned}
&\frac{\alpha !}{i !(\alpha-i)!}+9\frac{\alpha !}{(i-1) !(\alpha-i+1)!}=0 \\
\Leftrightarrow\ &\alpha-i+1+9i=0 \\
\Leftrightarrow\ &\alpha=-8i-1<0.\end{aligned}$$ Thus the coefficients of $x^{2\alpha+3-2i}y^{2i}$ in $R_{2}$ are not equal to $0$.

\(3) We have $$\begin{aligned}
R_{3}&=(x^{2}+3y^{2})^{2}(x^{2}-y^{2})^{\alpha} \\
 &=\sum_{i=0}^{\alpha+1} (-1)^{i} \left( \binom{\alpha}{i}-6\binom{\alpha}{i-1}+9\binom{\alpha}{i-2} \right)x^{2\alpha+4-2i}y^{2i}.\end{aligned}$$ If the coefficients of $x^{2\alpha+4-2i}y^{2i}$ in $R_{3}$ are equal to $0$, $$\binom{\alpha}{i}-6\binom{\alpha}{i-1}+9\binom{\alpha}{i-2}=0.$$ Then we have $$\begin{aligned}
&\frac{\alpha !}{i !(\alpha-i)!}-6\frac{\alpha !}{(i-1) !(\alpha-i+1)!}+9\frac{\alpha !}{(i-2) !(\alpha-i+2)!}=0 \\
\Leftrightarrow\ &(\alpha-i+2)(\alpha-i+1)-6i(\alpha-i+2)+9i(i-1)=0 \\
\Leftrightarrow\ &16i^{2}-(8\alpha+24)i+\alpha^{2}+3\alpha+2=0.\end{aligned}$$ We have $$i=\frac{4\alpha+12 \pm \sqrt{48\alpha+112}}{16}.$$ Since $i$ is an integer, $48\alpha+112$ is a square number.

Proof of Theorem \[thm:main upper bound III\] {#sec: proof 