My Math Forum (a^p)-(b^p)-(c^p) congruent to 0 (mod p) (with mild restrictive conditions)

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 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.

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