Question

If `a` and `b` are integers such that `(1/sqrt(2017))` is the solution of the quadratic equation `x^2+ax+b=0`, then `a+b` ?

Answer 1

There are no integers `a,b` that meet the condition.

Explanation:

Substitute `x=1/(sqrt(2017)` in the equation `x^2+ax+b=0` and you obtain
`1/2017 + a/sqrt(2017) + b =0`
`(1/2017+b) + a/sqrt(2017)=0`.

Since `1/sqrt(2017)` is irrational and `a, 1/2017+b` are rational, you have to satisfy:
`1/2017+b=0`, `a=0`
Solving the equation you find `(a,b)=(0,-1/2017)` and `b` is not an integer.