# Test Today....Quick Number Theory Question

by trap101
Tags: number, test, theory, todayquick
 P: 327 Let "a" be an odd integer. Prove that a2n (is congruent to) 1 (mod 2n+2) Attempt: By using induction: Base Case of 1 worked. IH: Assume a2k (is congruent to) 1 (mod 2k+2) this can also be written: a2k = 1 + (l) (2k+2) for some "l" IS: a2k+1 = a2k°2 = (a2k)2 Now I took 1 + (l) (2k+2) and substituted it into (a2k)2 and expanded: 1 + l ( 2k+3+ (l) 22k+4) is what I obtained after expanding and then simplifying it. But I know this isn't what I have to obtain when I try and show the K+1 case. What am I missing?
 P: 29 You've probably already had your test, but everything you have done so far is correct. Now you just need to show that 1 + l(2k + 3) + l2(22k + 4) is congruent to 1 (mod 2k + 3), which it is as far as I can tell...
P: 327
Nope. Test is 6pm my time, so I'm just tying up a few loose ends.

 Quote by Ansatz7 You've probably already had your test, but everything you have done so far is correct. Now you just need to show that 1 + l(2k + 3) + l2(22k + 4) is congruent to 1 (mod 2k + 3), which it is as far as I can tell...

How is : 1 + l(2k + 3) + l2(22k + 4) congruent to 1 (mod 2k + 3)?

i tried breaking it up: 1 + l(2k + 3) + l2(2k + 2)2......so it's this last term that's giving me a problem. I also had one other short question if you don't mind?

P: 29

## Test Today....Quick Number Theory Question

 Quote by trap101 i tried breaking it up: 1 + l(2k + 3) + l2(2k + 2)2......so it's this last term that's giving me a problem. I also had one other short question if you don't mind?
What is 22k+4 divided by 2k+3? Feel free to ask another question, though I can't guarantee that I'll be able to respond.
 P: 327 The little things....smh. Well this one is easier I think: Find the remainder when (17!(15) - (22)542)2343 divided by 19 Attempt: I started it like this: (let's just call this ≈ congruent for now (I don't know how to get it in this program) 1518 ≈ 1 (mod 19) (By Fermat's little) 17!(15)18 ≈ 1 (17!) (mod 19) => 17! ≈ 17! (mod 19) Also: 22 ≈ 3 (mod 19) => 22542 ≈ 3542 (mod 19) So altogether I have: (17! - 3542)2343 Now there is a factorial so I figure I'm going to have to use Wilson's Thm, but I can't see how to squeeze it in
P: 29
 Quote by trap101 The little things....smh. Well this one is easier I think: Find the remainder when (17!(15) - (22)542)2343 divided by 19 Attempt: I started it like this: (let's just call this ≈ congruent for now (I don't know how to get it in this program) 1518 ≈ 1 (mod 19) (By Fermat's little) 17!(15)18 ≈ 1 (17!) (mod 19) => 17! ≈ 17! (mod 19) Also: 22 ≈ 3 (mod 19) => 22542 ≈ 3542 (mod 19) So altogether I have: (17! - 3542)2343 Now there is a factorial so I figure I'm going to have to use Wilson's Thm, but I can't see how to squeeze it in
I wasn't familiar with Wilson's theorem (I last did number theory more than 2 years ago, and I haven't used it since) but from what I can see it tells you that 18! is congruent to -1 mod 19. 18! = 18 * 17!, so I believe you can use this to simplify 17!.
P: 327
 Quote by Ansatz7 I wasn't familiar with Wilson's theorem (I last did number theory more than 2 years ago, and I haven't used it since)
and here I was thinking that I might have some grandiose use for this in the future, maybe I'll figure some use for it. Thanks though, you've been a help.

 Related Discussions Engineering, Comp Sci, & Technology Homework 2 Introductory Physics Homework 2 Introductory Physics Homework 9 Calculus 5 Introductory Physics Homework 2