Prove by induction:
4n+5n+6n is divisible by 15 for positive odd integers
For n=2k−1,n≥1 (odd integer)
42k−1+52k−1+62k−1=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=2k−1 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=16⋅4k+25⋅5k+36⋅6k.
Subtracting the original expression, we get 15⋅4k+24⋅5k+35⋅6k. The first term is divisible by 15. Now note that
24⋅5k+35⋅6k=15⋅8⋅5k−1+15⋅14⋅6k−1
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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