is there an integer value of m such that the solution to (2m-1)x^2 - (2m + 1)x + 1 = 0 are also integers?

Question

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Answer 1

The discriminant is:

 

Δ = b² - 4ac = (2m+1)² - 4(2m-1) = 4m² - 4m + 5 = (2m - 1)² + 4

 

We need this to be a perfect square. Since 3² - 2² > 4, we know that Δ < 3² = 9

 

It is easy to then check that it must be Δ = 4. However, this only happens if 2m - 1 = 0, which makes m = 1/2 ∉ Z.

 

Thus there is no integer m for which the roots are integers.