Fermat's Little Theorem

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Network Security: Fermat's Little Theorem
Topics discussed:
1) Fermat’s Little Theorem – Statement and Explanation.
2) Solved examples to prove Fermat’s theorem holds true for a given set of values.
3) Solved examples to prove Fermat’s theorem does not hold true for a given set of values.

Music:
Axol x Alex Skrindo - You [NCS Release]

#NetworkSecurityByNeso #Cryptography #NetworkSecurity #FermatsLittleTheorem
  1. Neso Academy
  2. fermat's theorem
  3. fermat's little theorem
  4. fermat's little theorem questions
  5. fermat's little theorem solved examples
  6. fermat's little theorem solved problems
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Комментарии
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The simplest way to solve P=11, a=5
*We need to check P should always be a prime number ! *
Formula=] a^p-1 congruent (1modp)
5*11-1=(1mod11)
5*10=(1 mod11)
The rule for fermat Little problem is
The number which is in the left should always be larger than the number which is right of equal to sign

So in this case its not satisfying the given condition so we divide it using modular methods

The number 5 is subtracted, with p and written as -6 on the right hand side...

-6*10=1mod11
Now we split the power of ten
As (2*5)

(-6^2)(*5)=1mod11
36(*5)=1mod11
Now 36mod11=3

3(*5)=1mod 11
Mod of 3*5 mod 11= 1

Now 1=1mod 11

Its congruent

P=11 and a=5 proved !

rishith
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Fermat's theorem holds true for p=11, a=5

And thanks for this presentation!

shivamkandhare
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5^10 ≡ 1 mod 11
5^(2*5) ≡ 1 mod 11
3^5 ≡ 1 mod 11
243 ≡ 1 mod 11 (Valid)

Therefore, FLT holds true for p=11 and a=5.

rajeshprajapati
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NESO ACADEMY YOU ARE LITERALLYY A SAVIOUR!!!! THANKYOU SO MUCH.

sanika
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Given p=11, a=5
a^p-1 = 1(mod11)
5^11-1 =1(mod11)
5^10 =1(mod11)
-6^10 =1(mod11) {5-11= -6}
-6^5*2 =1(mod11) {6^5=7776 is ÷ by 11 and remainder is 10
10^2 =1(mod11)
100 =1(mod11) {100÷11 and remainder should be 1}

Ritesh_kumar
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12 is congurent to 1 (mod 11) which shows Fermat's theorem holds true for p=11, a=5
Thanks for the explaination:)

lakshitapatnaikuni
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please complete Database management systems fastly. I have in my current semester 😩

monicabattacharya
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Very great course thank you Neso Academy ❤

markuche
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Fermat's theorem holds true for p=11, a=5

thanmaijami
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Fermat little theorem holds for
a= 5 and p=11
By the Fermat little theorem
a^(p-1)=1(mod p)
5¹⁰=1(mod 11)
Let's check using the Euler phi totient function
Which states that a^phi(n)=1(mod n)
Provided that GCD of (a, n)=1
And in this case a=5 and n=11
Thus :::
5^phi(11)=1(mod 11)
Since the phi(11)=11-1=10
5¹⁰=1(mod 11)
Thus Fermat little theorem holds

danieldanmola
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nice explanation sir. continue like in this way

balajimetla
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p=11, a=2, It holds fermet's little theorem (homework question answer)

ANMOLKUMAR-ixdi
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can someone say how -2^(4*3) congruent 1 (mod 13)
changes to 3 ^ 3 congruent 1 (mod 13)

victorymindset-
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How to solve by using Fermat's theorem, if the question is 27^452 mod 113?

aathavang
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For home work question I guess
p=11 and a=5
so we are using a^p-1=1 mod p
5^11-1=1mod 11
5^10=1 mod 11
then 5 is smaller than 11 so we are reduce 11 from 5 is 6
-6^10=1 mod 11
-6^5*2 = 1 mod 11
here 6^5 = 7776
then divide by 11 so we will get remainder 10
10^2 = 1 mod 11
100 = 1 mod 11
here 100/ mod 11 = 1
so 1= 1
hence proved ..
if it's correct like this comment ..😊

BookwormUnited
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Fermat theorem true value for p=11 and a=5

shivambhardwaj
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P=15 and a=4 it's not holding false it's holding true

ajay.
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why did you take -2 in the place of 11 in the second example

KuladeepGompa
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Fermat's Theorem hold true for p=11 and a=5 sir

garunkumar
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is there a way to check my answer using a calculator ?

nasserkhamis