Question

please answer all the questions.

just rearranging. Explanation is not needed.

Use modular arithmetic to prove that 3|(221 – 1) for an integer n > 0. Hence, 3|(221 – 1) for n > 0. To show that 3|(221 – 1), we can show that (221 – 1) = 0 (mod 3). We have: (221 – 1) = (4” – 1) (mod 3) Then, (22n – 1) = (1 - 1) = 0 (mod 3) Since 4 = 1 (mod 3), then 4" = 1 (mod 3)

Answer 1

for n=7,37=2187 <7!= 5040 Suppore that for some k36, 32K! 4 3k+1 - 3.3k since k76, then KH>7> 3 Hence akti < (KH) - 3k t By 3 k <k! the induction hypothesis then okt 2 (kH) K! t okt <CKH)! Then t Then 3h <n! for all n76

n²Ln The equality holds for n=4 . 16524 Now suppone that that k < Kl for a < K1 for a given K34. cur goal is to show that (kt s (kH)! By the left-hand side: ek+ 12 = x² +2kti. the induction hypothesis we know that ck! Becaure K74 then Kol Therefore crtg² <K! + 2K+K t Therefore (Atg? {ki tak How 3ck then CKADE KIAKIK Therefore Ck tg² < KICKA) then (kto? <(kH)? t Then by Mathematical induction n² <n! for eve by every n34

Let Pon) be the statement that 3 n'tan) P() is true since P+201) = 3 is divisible d Assume that PCk) is true for some Ko 3 by This means 31 (K² + 24) We want to show that PCkH) is true. That is we want to show that al CKH²+20k+ 1) (kH) +20kH) = + 3 k²t 3 kt) + zkt2 CB+2k) + (3 k² 3 kt 3) the induction hypothesis I CEZ such that By 3 f2k=30 Therefore (kt +206+) = 3 ct 3 (K 42k+3) = 30 (+²+2K43) to then 3 | Ck Hitzekt) + Then by Mathematical induction By pom is true for every nyo

21) 4*+ 52015-) The Basir step is true =21 h for the inductive step assume that 21 divides 4k+'+ 5 2K-1 kle want to show that 2CKH)-1 21 | 4let? 320k KH 4.4 't s² s2k- (K42 szekant, Then 4 kt z zekn)-1 -2K- - 4 akt + 25 5² Then 4 it 20K+1)- 5 = 4 (4 kt) + (6 +21) 5² ㅏ Then Kt2 2CKt)-1 4 + 5 4 (41 321) +(21)54) ICH ER-1 4 +5 By the Induction hy pothesis 21 then 7 A E7 such that 4 + 5 21 A مل Therefore, 216K2+1)) A + 5 - 4(21A) +21.5 - 21 (4A+ s 2k-1) t 2 (841)-1 So 211 4 ts At1 ant 1 + 5 Then By mathematical Induction, 2114 whenever nis positive integer.

Let n be an integer not divisible by 5, then nal mods), n=2(mods) n 3 (mods) ar n = 4(mods) t If n = 1 mod 5) then nt1 = 14 = o mods If naz mods) then he 1 Z 241 = 15= omods If n = 3 (mods) then rt 1 = 34 -1 = 80= omods If n = 4 (mods) then not 1 4 -1 =25s=omods L; This implies n4, is divisible by s Then for every n not divisible ible bys, we have n4-Y = o mods)