How to prove that ∀n∈N,∃k∈Z:42n+1+3n+2=13⋅k
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 13⋅k and try to prove it via congruences. So I'd need to prove that 42n+1+3n+2≡0(mod13)⟶42n+3+3n+3≡0(mod13), that is, 42n+1+3n+2≡42n+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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