Monday, 11 April 2016

elementary number theory - Prove 4n+5n+6n is divisible by 15



Prove by induction:





4n+5n+6n is divisible by 15 for positive odd integers




For n=2k1,n1 (odd integer)



42k1+52k1+62k1=15N



To prove n=2k+1, (consecutive odd integer)



42k+1+52k+1+62k+1=(4)42k+(5)52k+(6)62k,




How do I substitute the statement where n=2k1 to the above, to factor out 15 in order to prove divisibility? Would it be easier to assume n=k is odd and prove n=k+2 is divisible by 15?


Answer



As you suggested, it's notationally simpler to suppose 4k+5k+6k is divisible by 15 and consider
4k+2+5k+2+6k+2=164k+255k+366k.


Subtracting the original expression, we get 154k+245k+356k. The first term is divisible by 15. Now note that
245k+356k=1585k1+15146k1

is likewise divisible by 15. Thus, 4k+2+5k+2+6k+2 is indeed divisible by 15.



Query: Where did we use that k is odd? Well, obviously to start the induction. But where else?



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