mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Blogorrhea > enzocreti

Reply
 
Thread Tools
Old 2022-09-13, 07:22   #1
enzocreti
 
Mar 2018

2×269 Posts
Default Proving every 6^(6+35j)=648 mod 23004

How to prove that every 6^(6+35j) for nonnegative j is congruent to 648 mod 23004?

By Fermat litle theorem I can say (71-1)=70, 35 divides 70, so 6^35=1 mod 71

But how to continue the proof?
enzocreti is offline   Reply With Quote
Old 2022-09-13, 07:55   #2
User140242
 
Jul 2022

5210 Posts
Default

Quote:
Originally Posted by enzocreti View Post
How to prove that every 6^(6+35j) for nonnegative j is congruent to 648 mod 23004?

By Fermat litle theorem I can say (71-1)=70, 35 divides 70, so 6^35=1 mod 71

But how to continue the proof?

Maybe this can help, you can simplify:

6^(6+35j) - 648 = 0 mod 23004

2*324 *(72*6^35j-1) = 0 mod (324*71)

2*324*71*6^35j+2*324*(6^35j-1) = 0 mod (324*71)

Last fiddled with by User140242 on 2022-09-13 at 07:57
User140242 is offline   Reply With Quote
Old 2022-09-13, 09:30   #3
enzocreti
 
Mar 2018

2×269 Posts
Default

Quote:
Originally Posted by User140242 View Post
Maybe this can help, you can simplify:

6^(6+35j) - 648 = 0 mod 23004

2*324 *(72*6^35j-1) = 0 mod (324*71)

2*324*71*6^35j+2*324*(6^35j-1) = 0 mod (324*71)


the congruence should be equal to

6^6=0 mod 4

6^6=0 mod 81

6^6=9 mod 71

so the least exponent k for which 6^k=1 is k=35 a divisor of (71-1)=0

so this should conclude the proof
enzocreti is offline   Reply With Quote
Old 2022-09-14, 00:32   #4
bbb120
 
"特朗普trump"
Feb 2019
朱晓丹没人草

22×3×11 Posts
Default

Quote:
Originally Posted by enzocreti View Post
How to prove that every 6^(6+35j) for nonnegative j is congruent to 648 mod 23004?

By Fermat litle theorem I can say (71-1)=70, 35 divides 70, so 6^35=1 mod 71

But how to continue the proof?
you can use Exhaustive method with the help of computer!
bbb120 is offline   Reply With Quote
Old 2022-09-14, 01:54   #5
ATH
Einyen
 
ATH's Avatar
 
Dec 2003
Denmark

23×7×61 Posts
Default

Quote:
Originally Posted by enzocreti View Post
How to prove that every 6^(6+35j) for nonnegative j is congruent to 648 mod 23004?
66+35j = 66 * 635j


Proof by induction:

Each time you increase j by 1, you multiply the previous result by 635 = 5184 (mod 23004)

Define: Sj = 66 * 635j (mod 23004)

S0 = 66 (mod 23004) = 648 (mod 23004)
S1 = S0 * 635 (mod 23004) = 648*5184 (mod 23004) = 648 (mod 23004)
Sj+1 = Sj * 635 (mod 23004) = 648 (mod 23004)
Q.E.D


It works because 648*635 is again 648 (mod 23004).
ATH is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Proving, for A > 1, (A+1)^n mod A = 0 MushNine Number Theory Discussion Group 9 2018-01-04 03:29
Primality proving of DB factors? jasonp FactorDB 3 2011-10-17 18:04
Primality proving CRGreathouse Software 13 2011-01-30 14:30
New Class of primes - proving algorithm AntonVrba Math 2 2008-10-06 00:53
Countdown to proving M(6972593) is #38 down to less than 10! eepiccolo Lounge 10 2003-02-03 05:15

All times are UTC. The time now is 01:18.


Wed Dec 7 01:18:55 UTC 2022 up 110 days, 22:47, 0 users, load averages: 0.81, 1.08, 1.09

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔