If (a + 1)power a, (where a is odd), then x – a =
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So you assume true up to an arbitrary integer k (the Inductive Hypothesis), then prove true for the k+1 case.
So consider dk≡1(mod2). So what happens when you multiply both sides by d? You get dk∗d≡d(mod2). As d is odd by the Inductive Hypothesis, we get that dk+1 is also odd.
Alternatively, you can look at this combinatorially as well. So
(2m+1)k=(2m+1)(2m+1)...(2m+1)=∏i=1k(2m+1)
By expanding, the last digit will always be 1. And every other term will be multiplied by 2m, which implies that every other term is even. We know the even numbers are closed under addition, we have an integer of the form
2x+1=∑i=0k(ki)(2m)i∗1k−i
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