Is there something preventing you from doing either part by force? Just put in every single value of and see what happens.
My book states the following:
If 'b' is any integer and the polynomial f(x)=(x^2)+bx+1 factors (poly mod 9), there exists 3 distinct non-negative integers 'q' less than than 9 so that f(q) = 0(mod 9).
How can this be proven?
If 'b' is any integer and the polynomial f(x)=(x^2)+bx+1 factors (poly mod 8), then f(x) is a square. E.g. f(x) = ((x+c)^2)(poly mod 8) where 0<c<8.
Can anyone think of values of b that makes this possible?
The question asks you to show that for all , if the polynomial factors, then it has a certain number of roots. Since we are working modulo 8 or 9, we can just check every single value of and see if the result is true. (There are only finitely many values.)
Are you having a problem with factoring or finding roots of polynomials?
Since you don't get what we're saying, let me try a different approach.
We're given factors modulo i.e.
We then get for some that has an inverse modulo .
Here's all with an inverse:
Looking at the table we see there are four cases to consider:
Thus if factors modulo then . Now check to see there are three roots for both cases (this should be very easy).