A difficult Maths questiona是一个三位数,b是一个一位数,且a/b,(a²+b²)/(ab+1)都是整数,求a+b的最大值与最小值

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AdifficultMathsquestiona是一个三位数,b是一个一位数,且a/b,(a²+b²)/(ab+1)都是整数,求a+b的最大值与最小值AdifficultMaths

A difficult Maths questiona是一个三位数,b是一个一位数,且a/b,(a²+b²)/(ab+1)都是整数,求a+b的最大值与最小值
A difficult Maths question
a是一个三位数,b是一个一位数,且a/b,(a²+b²)/(ab+1)都是整数,求a+b的最大值与最小值

A difficult Maths questiona是一个三位数,b是一个一位数,且a/b,(a²+b²)/(ab+1)都是整数,求a+b的最大值与最小值
设a/b=k(k为整数),
那么(a²+b²)/(ab+1)= (k²+1)b²/ (kb²+1) =整数
要使 (k²+1)b²/ (kb²+1) =整数
必然(k²+1)=(kb²+1) 或者 b²=(kb²+1) 或(k²+1)b=(kb²+1)(解之kb=1即a=b=k,显然不成立.或者b=(kb²+1) 显然不成立
四个等式中只有一个成立.也只有一个解.
b²=(kb²+1) b²=( ab+1)因为a是一个三位数,b是一个一位数,显然不成立.
(k²+1)=(kb²+1)解之k=b²
a=b.b.b即的三次方,因为a是一个三位数,b是一个一位数,b最大取9,最小取5,才能满足条件.
所以a+b的最大值729+9=738为与最小值是125+5=130

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