What is the Theorem regarding the number of polynomial zeros modulo p and H?

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The discussion centers on understanding a theorem in number theory regarding the number of polynomial zeros modulo a prime p and an irreducible polynomial H. It states that for a polynomial P of degree d, there can be at most d zeros that are pairwise not congruent modulo both p and H. The term "congruent mod p and H" refers to two polynomials having the same remainder when divided by H modulo p. An example is provided with specific polynomials and a prime, illustrating the congruence relationship. The conversation also seeks clarification on the definitions and implications of these congruences in the context of polynomial division.
Kontilera
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Hello I am 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.
 
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Kontilera said:
Then P has at most d polynomial zeros that are pairwise not congruent modulo p and H.
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.
 
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.
 
Kontilera said:
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.
You haven't said what H is.
Kontilera said:
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.
 
srry!
H is an non-constant polynomial whoose leading coeff is coprime to n.
 
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)}##?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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