Modular Arithmetic Proof with exponents

AI Thread Summary
The discussion focuses on proving the modular arithmetic statement (a+b)^p mod p = [(a^p mod p) + (b^p mod p)] mod p, where p is a prime number. A participant expresses confusion about how to approach the proof, questioning whether induction is appropriate. Another contributor suggests using the binomial expansion of (a+b)^p, highlighting that many binomial coefficients will be divisible by p. This approach is recommended over induction due to the nature of the coefficients in relation to prime numbers. The conversation emphasizes understanding the properties of binomial coefficients in modular arithmetic.
JPanthon
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Homework Statement



Let p be a prime number.
Prove:

(a+b)^p modp = [(a^p modp) + (b^p modp)]modp


Homework Equations



modular arithmetic.


The Attempt at a Solution



I honestly haven't the slightest clue.
Would induction be my best bet here?
If so, when I suppose the statement is true for (k+1), n isn't always prime anymore.

I used to be a biochem major and just switched into algebra, so I'm sorry if I seem retarded, I'm just very behind! Help please!
 
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JPanthon said:

Homework Statement



Let p be a prime number.
Prove:

(a+b)^p modp = [(a^p modp) + (b^p modp)]modp

Homework Equations



modular arithmetic.

The Attempt at a Solution



I honestly haven't the slightest clue.
Would induction be my best bet here?
If so, when I suppose the statement is true for (k+1), n isn't always prime anymore.

I used to be a biochem major and just switched into algebra, so I'm sorry if I seem retarded, I'm just very behind! Help please!

No, induction isn't your best bet for the very good reason you mention. Think about the binomial expansion of (a+b)^p.
 
Last edited:
Write the binomial expansion without knowing what p is. Just write it symbolically. Can you show many of the binomial coefficients are divisible by p?
 

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