# Number theory basics :)

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1. Feb 23, 2016

### Kontilera

Hello im currently learning some of the basics of number theory, and struggling to understand this Theorem. Could someone please explain it with maby a simple example? :)

THRM:(Number of polynomial zero mod p and H)
Let p be a prime number and let H be a polynomial that is irruducible modulo p. Furthermore let P be a polynomial that has degree d>=0 modulo p. Then P has at most d polynomial zeros that are pairwise not congruent modulo p and H.

2. Feb 24, 2016

### Staff: Mentor

What does the part in bold mean?
I understand what "congruent mod p" means, but I don't understand what "congruent mod p and H" means.

3. Feb 24, 2016

### Kontilera

okey, i will explain with an example. say P=X^2 and Q=2X^3+X-2, and p=3, dividing P by H we obtain a rest of +4.
Then dividing Q by H we obtain a rest of -5.
+4 is congurent to +1 mod 3, and -5 is congurent with +1 mod 3.
We now say that the polynomials P is congurent to Q mod(p,H). Because they have the same remainders when dividing by H mod n.

4. Feb 24, 2016

### Staff: Mentor

You haven't said what H is.

5. Feb 24, 2016

### Kontilera

srry!
H is an non-constant polynomial whoose leading coeff is coprime to n.

6. Feb 24, 2016

### Staff: Mentor

May I summarize:
We have a prime $p$ and polynomials $P(x), H(x) ∈ ℤ[x]$ where $\deg P = d$ and $H[x] \mod p$ is irreducible in $Z_p[x]$.
Then $d \mod p ≥ 0$. But this is always the case.
Now we have to show that $P(x)$ has at most $d$ zeros $\{x_1,...,x_d\}$ in $ℤ$ or in $ℤ_p$?
Or did you mean $\{ x-x_1,...,x-x_d \}$ as "polynomial zeros"?

Those are pairwise incongruent "modulo $(p,H)$" which you defined as follows:
Two polynomials $P(x),Q(x) ∈ ℤ[x]$ are congruent modulo $(p,H)$ if $\frac{P}{H} = \frac{Q}{H} \mod n$.
I suppose $n=p$? Or $n=d$? And the division of the polynomials is performed in which Ring? Or shall we divide $\frac{P(x_i)}{H(x_i)}$?