This document discusses polynomial congruences, which are expressions of the form f(x) ≡ 0 (mod m) where f(x) is a polynomial and m is the modulus. It provides several key points: 1) Polynomial congruences can be reduced by removing terms and reducing coefficients modulo m. 2) With a prime modulus p, every polynomial congruence is equivalent to one with degree less than p. 3) Roots of a polynomial congruence with a prime power modulus pk can be found recursively, starting with roots modulo p and extending the solutions modulo higher powers of p.

1. POLYNOMIAL
CONGRUENCE WITH
PRIME MODULI

2. Let 𝑓 𝑥 = 𝑎𝑖𝑥𝑖 + 𝑎𝑖−1𝑥𝑖−1 + ⋯ + 𝑎0 = 0 be an integral
polynomial. The largest integer 𝑘 suck that 𝑎𝑘 ≠ 0 is called the
degree of the polynomial, and the corresponding coefficient
called the leading term of the polynomial.
then the expression 𝒇 𝒙 ≡ 𝟎 𝒎𝒐𝒅 𝒎 is known as a
polynomial congruence

3. • if 𝑓 𝑥 = 𝑎𝑖𝑥𝑖 + 𝑎𝑖−1𝑥𝑖−1 + ⋯ + 𝑎0 , 𝑎𝑖 ≡ 𝑏𝑖 𝑚𝑜𝑑 𝑚 and 𝑔 𝑥 = 𝑏𝑖𝑥𝑖 + 𝑏𝑥𝑖−1 + ⋯ + 𝑏0
then clearly 𝒇 𝒙 ≡ 𝒈 𝒙 (𝒎𝒐𝒅 𝒎) for all 𝒙 . Hence, in a congruence 𝒇 𝒙 ≡
𝟎 𝒎𝒐𝒅 𝒎 we may reduce the coefficients modulo 𝒎, and in particular we may delete
terms 𝑎𝑖𝑥𝑖 with 𝑎𝑖 = 0 (mod 𝒎) without changing the solution set.
Example
𝟐𝟎𝒙𝟓
+ 𝟏𝟕𝒙𝟒
+ 12𝒙𝟐
+ 𝟏𝟏 (𝒎𝒐𝒅 𝟒)
is equivalent to the conguence
𝒙𝟒
+ 𝟑 ≡ 0 𝒎𝒐𝒅 𝟒
• And by trying -1, 0, 1, 2, we will find the solution x ≡ ±𝟏 𝒎𝒐𝒅 𝟒
𝑥4
+ 3 ≡ 0 𝑚𝑜𝑑 4
14
+ 3 ≡ 0 𝑚𝑜𝑑 4 −14
+ 3 ≡ 0 𝑚𝑜𝑑 4
1 + 3 ≡ 0 𝑚𝑜𝑑 4 1 + 3 ≡ 0 𝑚𝑜𝑑 4
4 ≡ 0 𝑚𝑜𝑑 4 4 ≡ 0 𝑚𝑜𝑑 4

4. • THEOREM 9.3
If p is a prime, then every polynomial congruence f(x) ≡ 0 𝑚𝑜𝑑 𝑝 is equivalent to
a polynomial congruence r(x) ≡ 0 𝑚𝑜𝑑 𝑝 , where r(x) is a polynomial with degree less than
p.
*Example
Consider the congruence 𝒙𝟏𝟏 + 𝟐𝒙𝟖 + 𝒙𝟓 + 𝟑𝒙𝟒 + 𝟒𝒙𝟑 + 𝟏 𝒎𝒐𝒅 𝟓
Division by 𝑥5 − 𝑥
𝑥11 + 2𝑥8 + 𝑥5 + 3𝑥4 + 4𝑥3 + 1 = (𝑥6 + 2𝑥3 + 𝑥2 + 1)(𝑥5 − 𝑥) + 5𝑥4 + 5𝑥3 + 𝑥 + 1
The given congruence is equivalent to the congruence 𝟓𝒙𝟒
+ 𝟓𝒙𝟑
+ 𝒙 + 𝟏 ≡ 0 𝑚𝑜𝑑 5
That is 𝑥 + 1 ≡ 0 𝑚𝑜𝑑 5
With sole solution 𝑥 ≡ 4 (mod 5)

5. • THEOREM 9.4 (LEMMA)
Assume 𝑛 ≥ 𝑝 and 𝑛 ≡ 𝑟 (𝑚𝑜𝑑 𝑝 − 1 , 𝑤ℎ𝑒𝑟𝑒 1 ≤ 𝑟 ≤ 𝑝 − 1. 𝑇ℎ𝑒𝑛 𝑥𝑛 ≡
𝑥𝑟 𝑚𝑜𝑑 𝑝 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥.
Example:
Consider the congruence 𝒙𝟏𝟏 + 𝟐𝒙𝟖 + 𝒙𝟓 + 𝟑𝒙𝟒 + 𝟒𝒙𝟑 + 𝟏 𝒎𝒐𝒅 𝟓
𝑝 − 1 = 5 − 1 = 4
11 ≡ 3, 8 ≡ 4, 5 ≡ 1 𝑚𝑜𝑑 4
Replace the terms 𝒙𝟏𝟏 , 𝟐𝒙𝟖 + 𝒙𝟓 𝒃𝒚 𝒙𝟑 + 𝟐𝒙𝟒 + 𝒙
This result in the polynomial 𝒙𝟑 + 𝟐𝒙𝟒 + 𝒙 +𝟑𝒙𝟒 + 𝟒𝒙𝟑 + 𝟏 = 𝟓𝒙𝟒 + 𝟓𝒙𝟑 + 𝒙 + 𝟏
𝟓𝒙𝟒 + 𝟓𝒙𝟑 + 𝒙 + 𝟏 ≡ 𝒙 + 𝟏 ≡ 0 𝒎𝒐𝒅 𝟓

6. A polynomial congruence with a general modulus may have more roots that the degree of the
polynomial
* For example, the congruence 𝑥2 − 1 ≡ 0 𝑚𝑜𝑑 8 ℎ𝑎𝑠 𝑡ℎ𝑒 𝑓𝑜𝑢𝑟 𝑟𝑜𝑜𝑡𝑠 1, 3, 5 𝑎𝑛𝑑 7.
However if the modulus is prime, then the number of roots can not exceed the degree unless all
coefficients of the polynomials are divisible by p.
Let p be a prime and let f(x) be an integral polynomial of degree n not all of whose coefficient
are divisible by p. Then the congruence f(x)≡ 0 𝑚𝑜𝑑 𝑝 ℎ𝑎𝑠 𝑎𝑡 𝑚𝑜𝑠𝑡 𝑛 𝑟𝑜𝑜𝑡𝑠.
Let p be a prime, and suppose that the polynomial f(x)has degree 𝑛 ≤ 𝑝 and leading coefficient
1. Use the division algorithm to write 𝑥𝑝
− 𝑥 = 𝑞 𝑥 𝑓 𝑥 + 𝑟 𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝑑𝑒𝑔 𝑟 𝑥 < 𝑑𝑒𝑔 𝑓 𝑥 .
Then f(x)≡ 0 𝑚𝑜𝑑 𝑝 has exactly 𝑛 roots if and only if every coefficient 𝑟 𝑥 is divisible by p.
Assume p be a prime and that d | (𝑝 − 1). Then the congruence 𝑥𝑑 − 1 ≡ 0 (𝑚𝑜𝑑 𝑝) has exactly d
roots.

7. POLYNOMIAL
CONGRUENCES WITH
PRIME POWER MODULI

8. • The general procedure for solving the polynomial
congruence 𝒇 𝒙 ≡ 𝟎 (𝒎𝒐𝒅 𝒎) when m is a prime power
𝒑𝒌
is to start with a root for the modulus p and use it to
generate a root( or in some cases several roots) modulo 𝒑𝟐
.
Using the same technique, we produce roots 𝒑𝟑
, 𝒑𝟒
, and so
on, until we finally obtain roots for the original modulus
𝒑𝒌
.

9. • The general procedure for finding all roots of 𝒇 𝒙 ≡ 𝟎 (𝒎𝒐𝒅 𝒑𝒌):
1. First find all solutions of the congruence 𝒇 𝒙 ≡ 𝟎 (𝒎𝒐𝒅 𝒑).
2. Select one, say 𝑎1;the there are either 0, 1 or p solutions of 𝒇 𝒙 ≡ 𝟎 𝒎𝒐𝒅 𝒑𝟐 congruent to
𝑎1 modulo p; if solution exist, they are found by solving the linear congruence f’(𝑎1)/p (mod
p). If there are no solutions, start again with a different 𝑎1.
3. If there are solutions of 𝒇 𝒙 ≡ 𝟎 𝒎𝒐𝒅 𝒑𝟐
, select one, say 𝑎2, and find the corresponding
roots of 𝒇 𝒙 ≡ 𝟎 𝒎𝒐𝒅 𝒑𝟑 by solving the congruence f’(𝑎2)t ≡ f’(𝑎2)/ 𝑝2 (mod p). Do this for
each root of f(x) ≡ 0 (𝑚𝑜𝑑 𝑝2).Note that since 𝑎2 ≡ 𝑎1 𝑚𝑜𝑑 𝑝 , f’(𝑎2) ≡ f’(𝑎1) (mod p), so we
do not need to calculate f’(𝑎2).
4. Proceeding in this fashion, we will eventually determine all solutions of 𝒇 𝒙 ≡ 𝟎 (𝒎𝒐𝒅 𝒑𝒌
)

10. Example : Solve the congruence 7𝑥6
+ 4𝑥 + 12 ≡ 0 𝑚𝑜𝑑 135
 Solution: Since 135 = 33. 5, the congruence is equivalent to the system
 7𝑥6 + 4𝑥 + 12 ≡ 0 𝑚𝑜𝑑 5
7𝑥6 + 4𝑥 + 12 ≡ 0 𝑚𝑜𝑑 33
A. 7𝑥6
+ 4𝑥 + 12 ≡ 0 𝑚𝑜𝑑 5 = 𝟐𝒙𝟐
+ 𝟒𝒙 + 𝟐 ≡ 𝟎 𝒎𝒐𝒅 𝟓 = (𝒙 + 𝟏)𝟐
≡ 𝟎 𝒎𝒐𝒅 𝟓
(𝑥 + 2)2
≡ 𝟎
𝒙 + 𝟏 ≡ 0 𝒙 ≡ −1 it has the sole root -1
7𝑥6
+ 4𝑥 + 12 ≡ 0 𝑚𝑜𝑑 3 = 𝑥2
+ 𝑥 ≡ 0 (𝑚𝑜𝑑 3)
* roots : 𝒙 ≡ 0 (𝑚𝑜𝑑 3) * 𝒙 ≡ −1 (𝑚𝑜𝑑 3)
7𝑥6
+ 4𝑥 + 12 ≡ 0 7𝑥6
+ 4𝑥 + 12 ≡ 0
7(0)6
+ 4 0 + 12 ≡ 0 (𝑚𝑜𝑑 3) 7(−1)6
+ 4 −1 + 12 ≡ 0 (𝑚𝑜𝑑 3) (𝑚𝑜𝑑 3)
12 ≡ 0 (mod 3) 15 ≡ 0 (𝑚𝑜𝑑 3)

11. B. We start from 𝑥1 = 0 and solve the linear congruence
* f’(𝑥1) = f’(𝑥1 /p (mod p) f’(0) = −
12
3
(𝑚𝑜𝑑 3) t ≡ −4 ≡ 2 (mod 3)
𝑥2 = 0 + 2 3 = 6 solves f(x) ≡ 0 (mod 𝟑𝟐
)
*f’(𝑥2) = f’(𝑥2 /p (mod p) f’(6) = −
6
9
(𝑚𝑜𝑑 3) t ≡ 2 (mod 3)
𝑥3 = 6 + 2 9 = 6 solves f(x) ≡ 0 (mod 𝟑𝟑
)
C. We start from 𝑦1 = −1 and solve the linear congruence
* f’(𝑦1) = f’(𝑦1 /p (mod p) f’(−1) = −
−15
3
(𝑚𝑜𝑑 3) t ≡ −5 ≡ 1 (mod 3)
𝑦2 = −1 + (1) 3 = 2 solves f(y) ≡ 0 (mod 𝟑𝟐
)
*f’(𝑦2) = f’(𝑦2 /p (mod p) f’(2) = −
2
9
(𝑚𝑜𝑑 3) t ≡ 2 (mod 3)
𝑦3 = 2 + (2) 9 = 20 solves f(y) ≡ 0 (mod 𝟑𝟑
)

12. D. To find the two solutions of our original congruence, we now use the Chinese
Remainder Theorem to solve the two systems
x ≡ −1 (mod 5) x ≡ −1 (mod 5)
x ≡ 24 (mod 33 ) and x ≡ 20 (mod 33 ).
The solutions are x ≡ 24 (mod 135) and x ≡ 74 (mod 135).