a 2 + b 2 = c 2. There are many proofs of this theorem. An elegant one only requires that we know that the area of a square of side L is L 2


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1 Pythgors Pythgors A right tringle, suh s shown in the figure elow, hs one 90 ngle. The long side of length is the hypotenuse. The short leg (or thetus) hs length, nd the long leg hs length. The theorem of Pythgors, whih is one of the most importnt in mthemtis euse of its mny pplitions, sttes tht = 2. There re mny proofs of this theorem. An elegnt one only requires tht we know tht the re of squre of side L is L = 2 Equl Ares Are = Are = +! Pythgoren triplets A fmous triplet is = 3, = 4, = 5 for whih = 5 2. Another one is = 5, = 12, = 13 for whih = Numer theory is the rnh of mthemtis tht dels with whole numers. Aove we sw two exmples of whole numer solutions of the eqution = 2. Are there infinitely mny solutions? The nswer is yes. One n onstrut infinitely
2 mny solutions y letting = m 2 n 2, = 2mn, nd = m 2 + n 2. For exmple, n = 1, m = 2 give = 3, = 4, = 5, n = 1, m = 4 give = 15, = 8, = 17, n = 2, m = 3 give = 5, = 12, = 13, n = 2, m = 5 give = 21, = 20, = 29, nd so on. Fermt s lst theorem A fmous question in numer theory is whether there re ny whole numer solutions to the eqution = 3. The nswer is no. In 1637 Pierre de Fermt stted without proof tht no whole numer triplets exist tht stisfy k + k = k where k is ny whole numer k = 2, 3, 4,... The est mthemtiins tried to prove this without suess until 1994 when it ws finlly proved y Andrew Wiles. Rtionl nd Irrtionl numers The integers, or whole numers, re...2,1,0,1,2,3... Rtionl numers re those tht n e written s the rtio or quotient of two whole numers, suh s 2/3, 3/2, 1/2, 5/1, 3/4,... The squre root of 2, 2 is numer tht when squred gives 2, ( 2 ) 2 = 2. The Pythgorens were upset to disover tht 2 is irrtionl, it nnot e written s the rtio of two whole numers. The proof is s follows. Let 2 = p q.
3 where p nd q, re two whole numers tht hve no ommon ftors. If initilly they hd ommon ftors, nel them. In prtiulr, p nd q re not oth even. I will show this is not possile. ( 2 ) 2 = 2 = ( ) 2 p = p2 q q 2, so p2 = 2q 2, therefore p 2 is even. Sine p 2 is even, we see tht p is even, so we n write p = 2m, where m is nother whole numer. We will show tht q is lso even. Sine p 2 = 2q 2 nd p = 2m, we get (2m) 2 = 4m 2 = 2q 2, or 2m 2 = q 2. But this mens tht q is lso even, whih is ontrdition. Prime Numers nd Pulikey Enription A prime numer (or prime) is nturl numer (1,2,3,4,5,...) tht hs extly two distint nturl numer divisors: 1 nd itself. The smllest twentyfive prime numers (ll the prime numers under 100) re: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Eulid proved tht there is n infinite numer of primes. RSA enryption (whih stnds for Rivest, Shmir nd Adlemn, three mthemtiins who first pulily desried it) involves puli key nd privte key. The puli key n e known to everyone nd is used for enrypting messges. Messges enrypted with the puli key n only e derypted using the privte key. The keys for the RSA lgorithm re generted the following wy (Wikipedi RSA Algorithm): (1) Choose two distint prime numers p nd q tht re kept seret.
4 For seurity purposes, the integers p nd q should e hosen t rndom, nd should e of similr itlength. Prime integers n e effiiently found using primlity test. (2) Compute n = pq. n is used s the modulus for oth the puli nd privte keys (3) Compute φ(n) = (p 1)(q 1). (4) Choose n integer e suh tht 1 < e < φ(n) nd the gretest ommon ftor of e nd φ(n) is 1. For exmple if e = nd φ(n) = 1 3 5, the gretest ommon ftor is 1. The puli key onsists of n nd the puli exponent e. Sometimes e = 3 is used. (5) determine d suh tht d e = 1modφ(n). This mens tht d e divided y φ(n) hs reminder of 1. For exmple with e = 8 nd φ(n) = 15, one needs d = 2. d is kept s the privte key exponent. The puli key onsists of the modulus n nd the puli (or enryption) exponent e. The privte key onsists of the privte p, q, nd exponent d whih must e kept seret. Alie trnsmits her puli key (n,e) to Bo nd keeps the privte key seret. Bo then wishes to send messge M to Alie. He first turns M into n integer 0 < m < n y using n greedupon reversile protool known s pdding sheme. A trivil one would e to just sustitute every hrter y its ASCII ode. He then omputes the iphertext orresponding to = m e (mod n). Bo then trnsmits to Alie. Alie n reover m from y using her privte key exponent d vi omputing m = d (mod n). Given m, she n reover the originl messge M y reversing the pdding sheme.
5 Fermt s Little Theorem The whole thing depends on Fermt s Little Theorem, whih sttes tht if p is prime numer, then for ny integer, p is multiple of p, whih in terms of modulr rithmeti sys p = (mod p). If is not multiple of p, then p 1 = 1(mod p)
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