Question 1036258
Let a,b be the zeroes of p(x).

{{{p(x) = x^2-p(x+1)-c  = x^2 - px - p - c}}} ==> -p = -(a+b) ==> p = a+b, 

and 

-p - c = ab.

Now (a+1)(b+1)= ab + a+b+1 =0 ==> -p-c + p + 1 = 0

==> c = 1, after simplifying.