Question

19．求不定方程2x⁴+x²y²+5y²=y⁴+10x²的全部整數(shù)解．

Answer 1

分析 先將方程化成（2x⁴+x²y²-y⁴）-5（2x²-y²）=0，再把左邊分解因式，從而得出（2x²-y²）=0或（x²+y²-5）=0，最后分別討論求解即可．

解答 解：原方程可化為：（2x⁴+x²y²-y⁴）-5（2x²-y²）=0，
∴（2x²-y²）（x²+y²-5）=0，
∴（2x²-y²）=0或（x²+y²-5）=0，
①當2x²-y²=0時，
∵y=±$\sqrt{2}$x，
∵不存在滿足此條件的整數(shù)x，y，
當x²+y²-5=0，即：x²+y²=5時，
∴$\left\{\begin{array}{l}{{x}^{2}=1}\\{{y}^{2}=4}\end{array}\right.$或$\left\{\begin{array}{l}{{x}^{2}=4}\\{{y}^{2}=1}\end{array}\right.$，
∴$\left\{\begin{array}{l}{{x}_{1}=1}\\{{y}_{1}=2}\end{array}\right.$或$\left\{\begin{array}{l}{{x}_{2}=1}\\{{y}_{2}=-2}\end{array}\right.$或$\left\{\begin{array}{l}{{x}_{3}=-1}\\{{y}_{3}=2}\end{array}\right.$或$\left\{\begin{array}{l}{{x}_{4}=-1}\\{{y}_{4}=-2}\end{array}\right.$或$\left\{\begin{array}{l}{{x}_{5}=2}\\{{y}_{5}=1}\end{array}\right.$或$\left\{\begin{array}{l}{{x}_{6}=2}\\{{y}_{6}=-1}\end{array}\right.$或$\left\{\begin{array}{l}{{x}_{7}=-2}\\{{y}_{7}=1}\end{array}\right.$或$\left\{\begin{array}{l}{{x}_{8}=-2}\\{{y}_{8}=-1}\end{array}\right.$．

點評 此題是非一次不定方程（組），主要考查了高次多項式的分解因式，完全平方數(shù)的特點，整數(shù)解的特點，解本題的關(guān)鍵是分解因式．