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# Daily Note #1 - Fermat's Little Theorem

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.

Note by Surya Prakash
1 year, 8 months ago

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Another 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 !! · 1 year, 8 months ago

Nice work, bro. Upvoted $$\ddot \smile$$. · 1 year, 8 months ago

Tks, i have checked your profile, it very impress, proud to be a vietnamese :D · 1 year, 8 months ago

Yep, proud to be Vietnamese :) · 1 year, 8 months ago