Modulus Congruence Proof

Date: 04/18/2001 at 03:11:48
From: Ooi Chin Wah
Subject: Number theory

Thanks for your answers to the two questions that I asked earlier. I 
would be very grateful if you can help me again. 

Prove 2^(3n+2)+21n = 4 mod (49)

Thanks.

Date: 04/18/2001 at 10:01:54
From: Doctor Rob
Subject: Re: Number theory

Thanks for writing back to Ask Dr. Math.

First let's verify the congruence modulo 7. We know 2^3 = 1 (mod 7). 
That implies that:

     2^(3*n+2) = 4*(2^3)^n = 4*1^n = 4 (mod 7)
     2^(3*n+2) + 21*n = 4 (mod 7)

and the congruence holds modulo 7. Now we know that:

         2^3 = 1 + 7
     2^(3*n) = (1+7)^n

Expand using the Binomial Theorem, and reduce modulo 49 = 7^2. Then 
multiply both sides by 4, and rearrange to get your answer.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/