Finding Integer Roots of $h+k=2016$

[Albert1]

given $h+k=2016$, and the two roots $\alpha \,\, and \,\, \beta $ of equation $x^2+hx+k=0$ are all integers , please find the value of:
$h,k,\alpha \,\, and \,\, \beta$

 

[kaliprasad]

given $h+k=2016$, and the two roots $\alpha \,\, and \,\, \beta $ of equation $x^2+hx+k=0$ are all integers , please find the value of:
$h,k,\alpha \,\, and \,\, \beta$

Spoiler

The roots are $\alpha,\beta$
$(x - \alpha) (x- \beta) = 0 $ or $x^2-(\alpha+ \beta) + \alpha\beta = 0$
comparing with given eqaution
$\alpha+ \beta= - h $ and $\alpha\beta= k$
from h + k = 2106 we get $\alpha\beta - (\alpha + \beta) = 2016 $
or $\alpha\beta - (\alpha + \beta) + 1 = 2017 $
or $(\alpha- 1)(\beta - 1) = 2017 $
as 2017 is a prime $\alpha = 2018$ and $\beta = 2$ and hence $h = - 2020,k = 4036$
or $\alpha = 2$ and $\beta = 2018$ and hence $h = - 2020,k = 4036$

 

[Greg]

Spoiler

$\alpha=2016,\,\beta=0,\,h=2016,\,k=0$ and any permutation thereof are also solutions.
Edit: sign error - the above is incorrect, sorry about that...

 

[kaliprasad]

Spoiler

$\alpha=2016,\,\beta=0,\,h=2016,\,k=0$ and any permutation thereof are also solutions.

Thanks Greg

Spoiler

forgot the product to be (-1) and (-2017) which gives the above solution subject to the restriction that permutation of
$\alpha,\beta$ is possible but not any permutation