Challenge problem: a + b + ab = p - 1

[lookagain]

This is my challenge problem to the community. It is based off of a problem here:

please find the answer with solution

If a and b are members of integers, find the values of a and b in the equation below: a + b + ab = 16

www.freemathhelp.com

Please show your work for the following.

The challenge problem:

Find all pairings (a, b), where a, b belong to the set of integers, and a and b are in terms of p,
where p represents a prime number.

\(\displaystyle a + b + ab \ = \ p - 1 \)

 

[lookagain]

These are some individual steps/hints, one at a time, of a possible method.
Or, you may choose your own route/method.

1)

Spoiler

a(b + 1) + b = p - 1

2)

Spoiler

a(b + 1) = -b + p - 1

3)

Spoiler

\(\displaystyle a \ = \ \dfrac{-b + p - 1}{b + 1} \)

4)

Spoiler

From long division:

\(\displaystyle a \ = \ -1 \ +\ \dfrac{p}{b + 1}\)

5)

Spoiler

The (integer) divisors of p are -p, -1, 1, and p.

I'll end the partial steps/hints here.

 

[nasi112]

I wish if I can understand what is going on here, but I am aware what prime numbers are

 

[Cubist]

I also like the method posted by @BigBeachBananas in the other thread

Spoiler

\(\displaystyle a + b + ab \ = \ p - 1 \)

\(\displaystyle a + b + ab + 1=p\)

\(\displaystyle (a + 1)(b + 1)= (1)(p)= (-1)(-p)=(p)(1)=(-p)(-1)\)