How many positive integers a less than 100 have a corresponding integer b divisible by 3 such that the roots of x^2-ax+b=0 are consecutive positive integers?

Question

Mathematics

Yasmin Larsen

27 November, 22:19

How many positive integers a less than 100 have a corresponding integer b divisible by 3 such that the roots of x^2-ax+b=0 are consecutive positive integers?

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Answers (1)

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Sierra Mendez · Answer 1

There are 24 such numbers

5, 7, 11, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 61, 65, 67, 71, 79, 83, 85, 91, 95

Step-by-step explanation:

Solving, we have

(x-y) (x - (y-1)) = x^2 - (2*y-1) x+y^2-y=x^2-ax+b comparison shows that

(2y-1) = a and y^2-y = b or

y (y-1) = b

This gives the following qualifying numbers

5, 7, 11, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 61, 65, 67, 71, 79, 83, 85, 91, 95