Question

Suppose $a$, $b$ and $c$ are integers such that the greatest common divisor of $x^2 ax b$ and $x^2 bx c$ is $x 1$ (in the set of polynomials in $x$ with integer coefficients), and the least common multiple of $x^2 ax b$ and $x^2 bx c$ is $x^3-4x^2 x 6$. Find $a b c$.

Answer 1

The GCD is
`x+1`, so
`x+1` divides both
`x^2+ax+b` and
`x^2+bx+c`. For some
`\alpha` and
`\beta` we can write

`x^2 + ax + b = (x + 1) (x - \alpha)`

`x^2 + bx + c = (x + 1) (x - \beta)`

Expanding the right sides, we get

`x^2 + ax + b = x^2 + (1 - \alpha) x - \alpha`

`x^2 + bx + c = x^2 + (1 - \beta) x - \beta`

`x+1` also divides the LCM, so

`x^3 - 4x^2 + x + 6 = (x + 1) (x - \alpha) (x - \beta)`

and expanding gives

`x^3 - 4x^2 + x + 6 = x^3 + (1 - \alpha - \beta) x^2 + (\alpha \beta - \alpha - \beta) x + \alpha \beta`

Matching up all the coefficients, we have

`\begin{cases} a = 1 - \alpha \\ b = -\alpha \\ b = 1 - \beta \\ c = -\beta \\ -4 = 1 - \alpha - \beta \\ 1 = \alpha \beta - \alpha - \beta \\ 6 = \alpha \beta \end{cases} \implies \alpha + \beta = 5`

All the unknowns are integers, and we have the prime factorization 6 = 2×3 that also satisfies 5 = 2 + 3.

• If
`\alpha=2` and
`\beta=3`, then
`a=-1`,
`b=-2`, and
`c=-3`.

• If
`\alpha=3` and
`\beta=2`, then there is no solution for
`a,b,c` since

`\begin{cases}b = -\alpha \\ b = 1-\beta \end{cases} \implies \alpha = \beta - 1`

but 3 ≠ 2 - 1 = 1.

It follows that
`a + b + c = -1 - 2 - 3 = \boxed{-6}`.