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

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.