So, starting with my first daily note, the topic is Fermat's Little Theorem.

**Topic: Number Theory**

Fermat's Little Theorem states that

**"For all natural numbers 'a' , \(a^{p} \equiv a \mod p\) , where 'p' is a prime number. "**

Let us prove this out.

Consider the binomial expansion for the prime 'p',

\( (a+b)^{p} = a^{p} + \binom{p}{1} a^{p-1} b +\binom{p}{2} a^{p-2} b^{2} + ....... + b^{p}\).

But since, \( p| \binom{p}{k} \forall k=1,2,3,.....p-1\). So, \( \binom{p}{1} a^{p-1} b +\binom{p}{2} a^{p-2} b^{2} + ....... + \binom{p}{p-1} a b^{p-1} = Multiple(p) \). This implies that, \( (a+b)^{p} \equiv a^{p} + b^{p} \mod p \).

Generalizing this we get, \( (a_{1}+a_{2} +.....a_{n})^{p} \equiv a_{1}^{p} + a_{2}^{p} +........ +a_{n}^{p} \mod p \). By taking \(a_{1}=a_{2}=.....=a_{n}=1\), we get \(n^{p} \equiv n \mod p\). That's it, we got the result.

Phewwwwwww!! We have proved it.

Fermat's Theorem is very useful in some problems based on Modular Arithmetic.

Now, if \((a,p) =1\),i.e. if \(a\) and \(p\) are coprime to each other, then \(a^{p-1} \equiv 1 \mod p\). This is known as Fermat's Little Theorem and it is a special case of Euler's Totient Theorem.

Now let us solve some problems.

*Problem 1(introductory):* Find the remainder when \(37^{1123}\) is divided by \(17\).

*Solution:* Observe that \(37 \equiv 3 \mod 17\) and from Fermat's Theorem \(37^{17} \equiv 37 \mod 17\). But \(1123 = 66 \times 17 + 1\). So, \(37^{1123} \equiv (37^{17})^{66} \times 37 \equiv 37^{66} \times 37 \equiv 37^{67} \equiv (37^{17})^{3} \times 37^{16}\) \( \equiv 37^{3} \times 37^{16} \equiv 37^{17} \times 37^{2} \equiv 37 \times 37^{2} \equiv 37^{3} \equiv 3^{3} \equiv 27 \equiv 10 \mod 17\).

So, \(37^{1123} \equiv 10 \mod 17\). This is can be even very easily using Fermat's littile theorem(*Try Yourselves*).

*Problem 2:* Find the remainder when \(2^{20} + 3^{30} + 4^{40} + 5^{50} + 6^{60}\) is divided by \(7\).

*Solution:* Observe that \((2,7)=(3,7)=(4,7)=(5,7)=(6,7)=1.\) So, \( 2^{20} \equiv 2^{2} \mod 7\), \(3^{30} \equiv 1 \mod 7\), \(4^{40} \equiv 4^{4} \equiv 4 \mod 7\), \(5^{50} \equiv 5^{2} \equiv 4 \mod 7\) and \(6^{60} \equiv 1 \mod 7\). So, \(2^{20} + 3^{30} + 4^{40} + 5^{50} + 6^{60} \equiv 4+1+4+4+1 \equiv 0 \mod 7\).

So, \(2^{20} + 3^{30} + 4^{40} + 5^{50} + 6^{60} \equiv 0 \mod 7\).

So, I think that I have given a clear picture on Fermat's Theorem. So, stay tuned for upcoming DAILY NOTES.

## Comments

Sort by:

TopNewestAnother way to prove Fermat's Theorem by using reduced residue system !!

Given that : { \( { r_{1},r_{2},...,r_{p-1} } \) } is the reduced residue system of p

Because of \((a,p) = 1\) so \((a.r_{i},p) = 1; (i = 1,2,...,p-1 ) \)

So that : \( a.r_{1}.a.r_{2}...a.r_{p-1} \equiv r_{1}.r_{2}...r_{p-1} \pmod{p}\)

\(\Rightarrow a^{p-1}.r_{1}.r_{2}...r_{p-1} \equiv r_{1}.r_{2}...r_{p-1} \pmod{p} \)

Thus,

\(a^{p-1} \equiv 1 \pmod{p} \)

The theorem is proved !! – Rony Phong · 1 year, 10 months ago

Log in to reply

– Trung Đặng Đoàn Đức · 1 year, 10 months ago

Nice work, bro. Upvoted \( \ddot \smile\).Log in to reply

– Rony Phong · 1 year, 10 months ago

Tks, i have checked your profile, it very impress, proud to be a vietnamese :DLog in to reply

– Trung Đặng Đoàn Đức · 1 year, 10 months ago

Yep, proud to be Vietnamese :)Log in to reply

Is there anything that you can add to the Fermat's Little Theorem Wiki page? – Calvin Lin Staff · 1 year, 10 months ago

Log in to reply