There are infinitely many solutions:

 where n is any integer positive negative or 0. 







Let 















It's easy to see that 1 is a solution
to the cubic polynomial in parentheses.
We use synthetic division

1|1 -1 -1  1
 |   1  0 -1
  1  0 -1  0

So we have factored to above as



And we can factor once more as



That has solutions

u=0, u=1, u=-1

Since 

, , 

We eliminate the third case because it will not give 
a real solution.

, 

The first gives any multiple of p or np,
where n is any integer, positive negative or zero.

The second gives ±p/2 plus any multiple of 2p,

or p/2 + 2kp = (1+2k)p/2.
where k is any integer, positive negative or zero. 

That's all odd multiples of p/2  

Since all multiples of p are also multiples
of ±p/2, all solutions are the multiples of
p/2 

np/2                       <--- answer
(where n is any integer, positive negative or zero.)

So the number of solutions is INFINITY!

Edwin