 My Math Forum (a^p)-(b^p)-(c^p) congruent to 0 (mod p) (with mild restrictive conditions)
 User Name Remember Me? Password

 Number Theory Number Theory Math Forum

 January 9th, 2015, 06:11 AM #1 Newbie   Joined: Jan 2015 From: Athens Posts: 4 Thanks: 0 (a^p)-(b^p)-(c^p) congruent to 0 (mod p) (with mild restrictive conditions) Hello there, I recently found out about a very interesting relationship in number theory: Given a = b + c where a, b and c are integers, it is true that: (a^p)-(b^p)-(c^p) is congruent to 0, modulo p, for any prime p. *Note that this relationship is a general version of Fermat's little theorem as: (a^p)-((a-1)^p)-1 is congruent to 0 modulo p So (a^p)-((a-2)^p)-(1^p)-1 is congruent to 0 modulo p (a^p)-((a-2)^p)-2 is congruent to 0 modulo p ... ... And so on until (a^p)-a is congruent to 0 modulo p Thereby probing Fermat's little theorem I have uploaded the proof for this, along with two other resulting lemmas on the following blog: Things Mathematical I would appreciate it if you: a) Read the proof and give me some feedback b) Let me know if you have knowledge of this relationship or a generalization of it. January 9th, 2015, 07:29 AM #2 Senior Member   Joined: May 2013 Posts: 118 Thanks: 10 Dear Stratos At a first glance,it seems to me that all of the three theorems are a direct consequence of Fermat's little theorem January 9th, 2015, 10:11 AM #3 Newbie   Joined: Jan 2015 From: Athens Posts: 4 Thanks: 0 Thanks for the observation As i think you can see through the proof, the derivation is completely unrelated to Fermat's Little theorem. However after your observation I realized that using Fermat's little theorem as a starting point: (a^p) = a (mod p) So assuming a = b + c (where b and c are both integers) We may say (a^p) = b + c (mod p) So it follows that (a^p) = b + c = (b^p) +(c^p) (mod p) And from that: (a^p) - (b^p) - (c^p) = 0 (mod p) That is an interesting observation, and thank you for pointing that out. It is interesting however, that the derivation that I have given does not involve Fermat's Little Theorem as a given, as it uses binomial expansion and a simple divisibility theorem. Tags apbpcp, conditions, congruence, congruent, fermat's little theorem, mild, mod, number theory, powers, prime, restrictive Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post claudiohorvi Number Theory 5 November 25th, 2013 06:15 AM coffee_leaf Number Theory 1 September 2nd, 2012 09:07 PM smslca Number Theory 4 January 28th, 2012 10:21 PM remeday86 Number Theory 7 July 2nd, 2010 07:12 AM sea_wave Number Theory 4 February 10th, 2010 11:16 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      