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January 9th, 2015, 06:11 AM   #1
Joined: Jan 2015
From: Athens

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(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.
ThingsMathematical is offline  
January 9th, 2015, 07:29 AM   #2
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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
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January 9th, 2015, 10:11 AM   #3
Joined: Jan 2015
From: Athens

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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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