Question

19．求不定方程2x⁴+x²y²+5y²=y⁴+10x²的全部整数解．

Answer 1

分析 先将方程化成（2x⁴+x²y²-y⁴）-5（2x²-y²）=0，再把左边分解因式，从而得出（2x²-y²）=0或（x²+y²-5）=0，最后分别讨论求解即可．

解答 解：原方程可化为：（2x⁴+x²y²-y⁴）-5（2x²-y²）=0，
∴（2x²-y²）（x²+y²-5）=0，
∴（2x²-y²）=0或（x²+y²-5）=0，
①当2x²-y²=0时，
∵y=±$\sqrt{2}$x，
∵不存在满足此条件的整数x，y，
当x²+y²-5=0，即：x²+y²=5时，
∴$\left\{\begin{array}{l}{{x}^{2}=1}\\{{y}^{2}=4}\end{array}\right.$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$\left\{\begin{array}{l}{{x}_{8}=-2}\\{{y}_{8}=-1}\end{array}\right.$．

点评 此题是非一次不定方程（组），主要考查了高次多项式的分解因式，完全平方数的特点，整数解的特点，解本题的关键是分解因式．