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October 3rd, 2015, 02:46 PM  #1 
Newbie Joined: Oct 2015 From: milton keynes Posts: 2 Thanks: 0  Impossible proof by induction
Prove (3^k ) + (7 ^ (k1)) + 8 is divisible by 12.
Last edited by skipjack; October 3rd, 2015 at 03:08 PM. 
October 3rd, 2015, 02:50 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra 
Why is this impossible?

October 3rd, 2015, 04:26 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,728 Thanks: 689 
True for k=1. Assume true for k, show true for k+1 (for k+1) $\displaystyle 3(3^k)+7(7^{k1})+8$ divisible by 12? Subtract assumption for k and get $\displaystyle 2(3^k)+6(7^{k1})$ divisible by 12? Expression is $\displaystyle 6(3^{k1}+7^{k1})$ Term inside parentheses is even! Therefore 6x term is divisible by 12! 

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