Wednesday, March 16, 2016

divisibility - Prove that 42n+1+3n+2:forallninmathbbN is a multiple of 13


How to prove that nN,kZ:42n+1+3n+2=13k



I've tried to do it by induction. For n=0 it's trivial.


Now for the general case, I decided to throw the idea of working with 13k and try to prove it via congruences. So I'd need to prove that 42n+1+3n+20(mod13)42n+3+3n+30(mod13), that is, 42n+1+3n+242n+3+3n+3(mod13)


But I have no clue how to do is. Any help?


Answer



42n+3+3n+3=16×42n+1+3×3n+2=13×42n+1+3×(42n+1+3n+2)


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