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   Author  Topic: Prove [4^(2n+1)] + [3^(n+2)] mod 13 =  0  (Read 405 times)
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Prove [4^(2n+1)] + [3^(n+2)] mod 13 =  0  
« on: May 10th, 2008, 4:24pm »		 Quote Quote Modify Modify
Show that [4^(2n+1)] + [3^(n+2)] is divisible by 13  n belongs to  Z+.
 
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Re: Prove [4^(2n+1)] + [3^(n+2)] mod 13 =  0  
« Reply #1 on: May 10th, 2008, 5:42pm »		 Quote Quote Modify Modify

Can be proved by induction
For n=1 F(n) = 91 which is divisible by 13
Let F(n) = 13k
Now for n+1 F(n+1) = 4^[2(n+1) +1] + 3^[(n+1] +2]
= 16*4^(2n+1) + 3*3^(n+2)
= 16 *[13k - 3^(n+2)] + 3*3^(n+2)
=16*13k - 13*3^(n+2)
 
and that expression is divisible by 13
 
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