每日一题：2020-06-04

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每日一题: 2020-06-04

题目: 设方程x2−px+q=0,x2−qx+p=0x^2-px+q=0, x^2-qx+p=0 的根都是正整数, 求正整数p,qp,q 的值.

参考思路

设方程x2−px+q=0x^2-px+q=0 的两根为x1,x2x_1,x_2; 方程x2−qx+p=0x^2-qx+p=0 的两根为x3,x4x_3,x_4.
由韦达定理, 有x1+x2=p,x1x2=q,x3+x4=q,x3x4=px_1+x_2=p, x_1x_2=q, x_3+x_4=q, x_3x_4=p
所以x1+x2=x3x4,x1x2=x3+x4⇒x1x2−x1−x2+x3x4−x3−x4=0x_1+x_2=x_3x_4, x_1x_2=x_3+x_4\Rightarrow x_1x_2-x_1-x_2+x_3x_4-x_3-x_4=0
因此有(x1−1)(x2−1)+(x3−1)(x4−1)=2(x_1-1)(x_2-1)+(x_3-1)(x_4-1)=2, 所以(x1−1)(x2−1)=2,1,0(x_1-1)(x_2-1)=2,1,0 时,
(x3−1)(x4−1)(x_3-1)(x_4-1) 相应取0,1,20,1,2.
\[
\left\{\begin{array}{lr} (x_1-1)(x_2-1)=2 \\ (x_3-1)(x_4-1)=0 \end{array}\right.\Rightarrow \left\{\begin{array}{lr} p=5 \\ q=6 \end{array}\right.
\]
同理可求得另外两组解, 综上
\[
\left\{\begin{array}{lr} p=5 \\ q=6 \end{array}\right., \left\{\begin{array}{lr} p=4 \\ q=4 \end{array}\right., \left\{\begin{array}{lr} p=6 \\ q=5 \end{array}\right.
\]