My solution is a little involved
I call as 'a' and as 'b' to avoid TeX.
First of all ,if a and b are integers, so are a-1 and b-1.
By part {i}, they satisfy '
So if a-1 and b-1 satisfy
then (a-1)(b-1) = p.
Since p is prime,the only possible solutions are
(a,b) = (2,p+1),(p+1,2),(0,-p-1),(-p-1,0)
We first claim that the first two solutions are invalid.
If either of a or b were 2, then they satisfy x^2 + kx + (p-k-1) = 0
that means
But this is not possible since p>0 and k>0 .
So look at the remaining solutions,
Now that means either of a or b is 0.
Substitute in x^2 + kx + (p-k-1) = 0
And get k = p-1
All done again