計算第1題
\(\begin{align}
  & {{a}^{2}}{{b}^{2}}=4{{a}^{5}}+{{b}^{3}} \\
& {{a}^{2}}\left( {{b}^{2}}-4{{a}^{3}} \right)={{b}^{3}} \\
\end{align}\)
令\(a=md,b=nd,\left( a,b \right)=d\)
\(\begin{align}
  & {{m}^{2}}{{d}^{2}}\left( {{n}^{2}}{{d}^{2}}-4{{m}^{3}}{{d}^{3}} \right)={{n}^{3}}{{d}^{3}} \\
& {{m}^{2}}\left( {{n}^{2}}d-4{{m}^{3}}{{d}^{2}} \right)={{n}^{3}} \\
& m=1,a=d \\
& {{n}^{2}}a-4{{a}^{2}}={{n}^{3}} \\
& 4{{a}^{2}}-{{n}^{2}}a+{{n}^{3}}=0 \\
& a=\frac{{{n}^{2}}\pm n\sqrt{{{n}^{2}}-16n}}{8} \\
\end{align}\)

令\({{n}^{2}}-16n={{k}^{2}}\quad ,k\in N\)
\(\begin{align}
  & {{\left( n-8 \right)}^{2}}-{{k}^{2}}=64 \\
& \left( n-8+k \right)\left( n-8-k \right)=64 \\
& \left( n-8+k,n-8-k \right)=\left( 32,2 \right),\left( 16,4 \right),\left( 8,8 \right),\left( -2,-32 \right),\left( -4,-16 \right),\left( -8,-8 \right) \\
& \left( n,k \right)=\left( 25,15 \right),\left( 18,6 \right),\left( 16,0 \right),\left( -9,15 \right),\left( -2,6 \right) \\
&  \\
& a=\frac{{{n}^{2}}\pm nk}{8}=\frac{n\left( n\pm k \right)}{8} \\
& \left( n,k \right)=\left( 25,15 \right),\left( a,b \right)=\left( 125,3125 \right) \\
& \left( n,k \right)=\left( 18,6 \right),\left( a,b \right)=\left( 54,972 \right),\left( 27,486 \right) \\
& \left( n,k \right)=\left( 16,0 \right),\left( a,b \right)=\left( 32,512 \right) \\
& \left( n,k \right)=\left( -9,15 \right),\left( a,b \right)=\left( 27,-243 \right) \\
& \left( n,k \right)=\left( -2,6 \right),\left( a,b \right)=\left( -1,2 \right),\left( 2,-4 \right) \\
\end{align}\)
共7組解

填充第2題
\(\begin{align}
  & {{\left( 3x-2 \right)}^{2}}<a{{x}^{2}} \\
& \left( 9-a \right){{x}^{2}}-12x+4<0 \\
\end{align}\)
恰有2個整數解，易知\(a\ne 9\)

令二次函數\(f\left( x \right)=\left( 9-a \right){{x}^{2}}-12x+4\)
若\(a>9,f\left( x \right)\)圖形開口朝下，\(f\left( x \right)<0\)會有無限多個整數解
故\(a<9,f\left( x \right)\)圖形開口朝上，判別式\({{\left( -12 \right)}^{2}}-4\left( 9-a \right)\times 4>0\quad \to \quad a>0\)
\(0<a<9\)

若\(f\left( 1 \right)=1-a>0\quad \to \quad a<1\)，頂點橫坐標\(=-\frac{-12}{2\left( 9-a \right)}=\frac{6}{9-a}<\frac{3}{4}\)，此時\(f\left( x \right)<0\)無整數解
故\(f\left( 1 \right)<0\)且\(f\left( x \right)<0\)的二個整數解為1和2
\(\begin{align}
  & f\left( 2 \right)=-4a+16<0\quad \to \quad a>4 \\
& f\left( 3 \right)=-9a+49\ge 0\quad \to \quad a\le \frac{49}{9} \\
& 4<a\le \frac{49}{9} \\
\end{align}\)

填充第 5 題
這種題目考填充，會不會有人畫一畫再用量角器量啊？還是連量角器都不能帶進去？

用三角函數可以湊出答案是 10 度

不過還是等高手來個漂亮的純幾何解吧！

應該都正確，第 4 題是 2006 AIME 的題目

這樣的話，一次項係數是負的，應該不是題目要的

剛算了一下，如果各項係數要正整數或 0，答案是無解

對!

m 和 n 互質

計算第二題
另解