The document discusses Fermat's theorem and its corollary related to sets of residues modulo m, alongside Euler's theorem as a generalization of Fermat's theorem. It provides a detailed proof of these theorems and offers examples, including a practical application to determine the last digit of 3400 through modulo calculations. The content is targeted towards students of mathematics to enhance their understanding of these number theory concepts.

1. Euler-Fermat theorem
Dr. Kuparala Venkata
Vidyasagar
Lecturer in Mathematics
SVLNS Govt. Degree
College,
Bheemunipatnam

2. LEARNING OUTCOMES
Fermat’s theorem
Corollary ON Fermat’s theorem
Set of residues modulo 𝑚
Reduced set of residues modulo 𝑚
Theorems based on residue modulo m
Euler's theorem or Euler's generalization of
Fermat's theorem
Fermat's theorem from Euler's theorem
Examples

3. Fermat’s theorem
Statement: If. 𝑃 is a prime and (𝑎, 𝑝) = 1 then 𝑎𝑝−1 ≡
1 mod𝑝
proof. Given that 𝑝 is a prime and (𝑎, 𝑝) = 1
claim: 𝑎𝑝−1
≡ 1(mod𝑝)
Since (𝑎, 𝑝) = 1 By known theorem
When the numbers 𝑎, 2𝑎, 3𝑎 … (𝑝 − 1)𝑎 are derided by p,
remainders are 1,2,3,….. p-1 not necessarily in this order
𝐿𝑒𝑡 𝑎 ≡ 𝑟1(mod𝑝)
2𝑎 ≡ 𝑟2(mod𝑝)
(𝑝 − 1)𝑎 ≡ 𝑟𝑝−1(mod𝑝)
But 𝛾1, 𝛾2; 𝑟𝑝−1 are also remainders obtained by dividing
𝑎, 2𝑎, ⋯ (𝑝 − 1)𝑎 by 𝑝
∴ 𝛾1, 𝑟2, ⋯ 𝑟𝑃−1 ≡ 1 ⋅ 2 ⋅ 3 ⋯ (𝑃 − 1) ⟶ (1)
From the above congruent relations

4. But 𝛾1, 𝛾2; 𝑟𝑝−1 are also remainders obtained by dividing
𝑎, 2𝑎, ⋯ (𝑝 − 1)𝑎 by 𝑝
∴ 𝛾1, 𝑟2, ⋯ 𝑟𝑃−1 ≡ 1 ⋅ 2 ⋅ 3 ⋯ (𝑃 − 1) ⟶ (1)
From the above congruent relations
(𝑃 − 1)! 𝑑𝑝−1 ≡ (𝑝 − 1)! (modp)
Since, 𝑃 is prime
(𝑃, 1) = 1, (𝑃, 2) = 1 … (𝑃, 𝑃 − 1) = 1
⇒ (𝑃, (𝑃 − 1)!) = 1
𝑎−1 ≡ 1(mod𝑝)
Hence the proof

5. If 𝑃 is a prime and a is any integer then
𝑎𝑝
≡ 𝑎(mod𝑝)
Case (i). Let (𝑎, 𝑝) ≠ 1
⇒ 𝑝 ∣ 𝑎 and
𝑎 ≡ 0 mod𝑝
ℎ𝑒𝑛𝑐𝑒 𝑎𝑃
≡ 0(mod𝑝)
and 0 ≡ 𝑎(mod𝑝)
By Transitive property
𝑎𝑝
≡ 𝑎 𝑚𝑜𝑑 𝑝
case(ii):
Let (𝑎, 𝑝) = 1
By Fermat's theorem
𝑎𝑝−1
≡ 1(mod𝑝)
𝑎𝑝
⋅
1
𝑎
≡ 1(mod𝑝)
𝑎𝑝
≡ 𝑎(mod𝑝)
Hence the proof
→ If 𝑃 is prime and (𝑎, 𝑝) = 1 then 𝑎𝑝−1 − 1 = 𝑀(𝑃)

6. Definition1.: A Set of integers 𝑥0, 𝑥1, 𝑥2, ⋯ 𝑥𝑚−1 is
called a set of residues modulo 𝑚 if each 𝑥𝑖(𝑖 =
0,1,2, ⋯ 𝑚 − 1) belongs to one and only one residue class
modulo.
Definition2.: A Set of integers 𝑥1, 𝑥2 … , 𝑥𝜙𝑚) is called a
reduced set of residues modulo 𝑚 if exactly one of them,
lies in each, residue class relatively prime to 𝑚,
Eg: {1,5} is a reduced set of residues modulo 6
= {0,1,2,3,4,5}
(∵ (1,6) = 1, (5; 6) = 1)
→ 𝜙(𝑚) is the number of elements in every reduced Set
of residues modulo 𝑚.

7. THEOREM: If 𝑟1, 𝑟2, … 𝑟𝜙(𝑚), is a set of reduced residue modulo 𝑚
and (𝑎, 𝑚) = 1 then 𝑎𝑟1, 𝑎𝑟2, … 𝑎𝑟𝜙(𝑚) is also a set of reduced
residue modulo 𝑚.
Given that 𝑟1, 𝑟2, … 𝑟𝜙(𝑚) is a set of
reduced residue modulo 𝑚 and
(𝑎, 𝑚) = 1
The
set 𝑎𝑟1, 𝑎𝑟2, … 𝑎𝑟𝜙(𝑚,} ℎ𝑎𝑠 𝜙(𝑚)
elements
Since 𝑟1, 𝑟2, … 𝑟𝜙(𝑚}, is a reduced
Set of residues modulo 𝑚.
⇒ Each 𝛾𝑖(𝑖 = 1,2, … − 𝜙(𝑚)) is
relatively prime to 𝑚
i.e. 𝑟𝑖, 𝑚 = 1 for 𝑖 = 1,2 … 𝜙(𝑚)
Since 𝑎, 𝑚 = 1, 𝑟𝑖, 𝑚 = 1
⇒ 𝑎𝑟𝑖
, 𝑚 = 1 for 𝑖 = 1,2, … 𝜙(𝑚)
∴ Each a𝑟𝑖 is relatively prime to 𝑚
Suppose two of the numbers
𝑎𝑟𝑖
, 𝑎𝑟𝑗 are Congruent
𝑎𝑟𝑖 ≡ 𝑎𝑟
𝑗(mod𝑚)
⇒ 𝑟𝑖 ≡ 𝑟
𝑗(mod𝑚) (∵ (𝑎, 𝑚) = 1)
This is impossible
∵ 𝛾𝑖, 𝛾𝑗 are two members of
reduced Set of residues modulo m
{they are incongruent)
no two elements a𝑟𝑖, a𝑟
𝑗 are
congruent
Hence 𝑎𝑟1, 𝑎𝑟2, … ar 𝜙(𝑛1 is also reduced set of residues modulo m.

8. Euler's theorem (Euler's generalization of Fermat's theorem)
Statement: If (𝑎, 𝑚) = 1 then 𝑎 ≡
𝜙(𝑚)
1(mod𝑚)
Proof: Given that (𝑎, 𝑚) = 1
Claim: 𝑎 ≡
𝜙(𝑚)
1(mod𝑚)
Let 𝑟1, 𝑟2, … 𝛾𝜙(𝑚,𝑗 be a
reduced set of
residues modulo 𝑚
Since(𝑎, 𝑚) = 1,
By known theorem
𝑎𝑟, 𝑎𝑟2, … ar𝜙(𝑚}
is also reduced set
of residues modulo 𝑚.
∴ Each a𝑟i is congruent modulo 𝑚
to one and only one 𝛾𝑗
𝐿𝑒𝑡 𝑎𝑟1
≡ 𝑥1(modm)
𝑎𝑟2 ≡ 𝑥2(modm)
−−−− −
𝑎𝑟𝜙(𝑚) ≡ 𝑥𝜙(𝑚)(modm)
Multiplying the above congruence
𝑎𝑟1, 𝑎𝑟2 … 𝑎𝑟𝜙(𝑚) ≡ 𝑥1 ⋅ 𝑥2 ⋯ 𝑥𝜙(𝑚)(modm)
⇒ 𝑟1 ⋅ 𝑟2 ⋯ 𝑟𝜙 𝑚 ⋅ a^𝜙(𝑚) ≡ 𝑥1 ⋅ 𝑥2 … 𝑥𝜙(𝑚(modm)
⇒ 𝛾1 ⋅ 𝛾2 … 𝛾∅ 𝑚 a^𝜙(𝑚) ≡ 𝛾1𝛾2 … 𝛾𝜙(𝑚)(modm)
∵ 𝑟1, 𝑟2; 𝑟𝜙(𝑚), are precisely numbers 𝑥1 ⋅ 𝑥2 … 𝑥𝜙(𝑚)
⇒ a ≡
𝜙(𝑚)
1(mod 𝑚)
∵ 𝑟1, 𝑟2 … 𝑟𝜙 𝑚 , 1 = 1
Hence the proof

9. Derive Fermat's theorem from Euler's
theorem
corollary: If 𝑚 = 𝑝 and 𝑎 ≡
𝜙(𝑚)
(mod𝑝) Then
𝑎𝑝−1
≡ 1(mod𝑝)
proof:
If 𝑚 = 𝑝 is a prime
𝜙(𝑝) = 𝑝 − 1
Since 𝑎
𝜙(𝑚)
≡ 1(modp)(𝐵𝑦 Euler's theorem)
⇒ 𝑎𝑝−1
≡ 1(mod𝑝)
Hence the proof

10. Question: What is the last digit in the ordinary decimal
representation of 3400
?
Solution: The last digit in the ordinary decimal representation of
3400
is the remainders when 3400
is divided by 10
By division algorithm 3400 ≡ 10q + 𝑟 where q, 𝑟 ∈ 𝑍 and 0 ⩽
𝑟 < 10
∴ 𝑟 is the last digit such that
3400 = 10q + 𝑟
3400
≡ 𝑟(mod10)
We have to find 𝑟 such that
3400 ≡ 𝑟(mod10)
Since 5 is prime (3,5) = 1
By Fermat's theorem
35−1
≡ 1(mod5)
34 ≡ 1(mod5) − (1)
Since 2 is prime and (3,2) = 1
By fermat' theorem
32−1
≡ 1(mod2)
31 ≡ 1(mod2)
34 ≡ 14(mod2) − (2)
From (1) & (2) , 34
≡ 1 (mod of
L.C.M of 5,2 )
)
34
≡ 1(mod 10
34 100
≡ 1100
(mod10)
3400
≡ 1(mod10)
𝑟 = 1
Hence the last digit =1.