Question

已知a.b为正整数，试问关于x的方程x^2

已知a.b为正整数，试问关于x的方程x^2-abx+1/2(a+b)是否有两个正整数解？如果有请把它证出来；如果没有，请给予证明。

Answer 1

有唯一可能的整数解，理由如下：
设x1,x2为可能满足的解
那么x1+x2=ab x1*x2=1/2(a+b)
显然a+b必须为偶数 那么设a=k-t b=k+t (k>t)
那么x1+x2=k^2-t^2 x1x2=k (1)
由于x1>=1 x2>=1 所以1/x1+1/x2=(x1+x2)/(x1x2)<=2
也即k^2-t^2<=2k 并由k>t知 t<k<=1+√(1+t^2) （2）
显然t<√(1+t^2)<t+1 那么t+1<1+√(1+t^2)<t+2
那么由不等式(2)知k=t+1 代回(1)式
得x1+x2=2t+1 x1x2=t+1
另外 由于(x1-x2)^2必须为整数 即
(x1-x2)^2=(x1+x2)^2-4x1x2=4t^2-3 则4t^2-3必须为完全平方数
设4t^2-3=p^2 也即(2t-p)(2t+p)=3=1*3
那么2t-p=1 2t+p=3 解得t=1
所以将t=1 k=t+1 带回可知a=1 b=3
整数解为x1=1 x2=2
综上 当且仅当a=1 b=3时方程有两个整数解x1=1 x2=2

Answer 2

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