I'm having trouble with this one, can anyone give me any hints?
Prove that if is any integer and the polynomial factors (poly mod 9), then there are three distinct non-negative integers less than 9 such that .
If factors then we can write . Now to have we want . We can have as one solution. We also have as a second solution. But we also have as a third solution.
Thus, there are at most three solutions. There are not necessarily exactly three solutions. Just consider as a conterexample.
Hi
Because 0=9=1*9 (which is equivalent to saying that one of the factors is 0, since 0=9) and also 0=9=3*3.
True.But isin't there still three solutions if ?
i.e., ,
.
Like TPH said, if it can be factored, then f(x)=(x+b)(x+c). And we knew from above that there are defined solutions (if we're restricted to the modulus) :
- b=3 mod 9 and c=3 mod 9 : this makes one solution, because b & c have the same modulus
- b=0 mod 9 and then c=1 mod 9 : two solutions
therefore, three distinct solutions