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Today’s Class is about Modular arithmetic. Ok modular arithmetic only works with reminder when some number is divided by another. 11 +2^2 +3^3 +... 20202020 ≡ 1+0+1+0+· · · ≡ 1010 ≡ 0 (mod 2) So the sum is even. The main idea is 3^3 ≡ −1 (mod 7) 32014 ≡ 3 2013+1^ ≡ 32013_._ 3 ≡ (3^3 )^671 ∗ 3 = (−1)^671 ∗ 3 = − 1 ∗ 3 = −3 = 4 (mod 7) The main idea is 3^2 ≡ −1 (mod 10) 381 = 3^80 ∗ 3 = (3^2 )^40 ∗ 3 = (−1)^40 ∗ 3 = 1 ∗ 3 = 3 (mod 10) 3 n^ = 3 (mod 10) if n = 4 k + 1 p > 3 and prime so p = 1 or 2 (mod 3) if p = 1 (mod 3) then p + 2 = 0 (mod 3)
p = 2 (mod 3) then p + 4 = 0 (mod 3) p = 3 is a solution. as p > 3 we know that p is odd let p = 2 k +1. now 2 p^ ≡ 22 k +1^ = 2^2 k^ ∗ 2 = 4 k^ ∗ 2 = 1 k^ ∗ 2 = 1 ∗ 2 = 2 (mod 3) as p > 3 and prime. p = 1 or 2 (mod 3) p^2 = 1 (mod 3) 2 p^ + p^2 = 2 + 1 = 0 (mod 3) x^2 , y^2 = 0 or 1 (mod 4) ∴ x^2 + y^2 = 0 , 1 or 2 (mod 4) But 2019 = 3 (mod 4).
(mod p ) We proved that, ap^ = a (mod p ) Now if ( a, p ) = 1, Then we can divide :) ap −^1 = 1 (mod p ) 27 −^1 = 2^6 = 1 (mod 7) check that , 2011 is a prime. and (2011 , 2010)
20102011 −^1 = 2010^2010 = 1 (mod 2011) 20102010 − 1 = 0 (mod 2011) If p | a or p | b Then p | ab ( bp −^1 − ap −^1 ) as p divides at least one of a, b. Now if p ∤ a and p ∤ b then ( a, p ) = ( b, p ) = 1. Then by Fermat’s little theorem , bp −^1 =
ap −^1 = 1 (mod p ) Then ( bp −^1 − ap −^1 ) = 0 (mod p )
