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I've came across this post:

Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$.

and the very last comment suggests to prove the above statement through

$2^=(3−1)^ =3+(−1)^$

I don't quite understand how one can conclude the last part. What I thought so far is that 3k is standing obviously for a number, which is divisible by 3 - however I don't see where this assumption comes from and what I am most interested in, how come the part with $(−1)^$?

Thank you very much for any help.

Greta
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1 Answers1

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You can use the binomial theorem $$(x+y)^n=\sum_{t=0}^n{n \choose t}x^ky^{n-k}$$ to show this. We have

$$\begin{array}{}2^n\\&=(3-1)^n\\&=\sum_{t=0}^n{n \choose t}\cdot 3^t\cdot(-1)^{n-t}\\&={n \choose 0}\cdot 3^0\cdot(-1)^{n-0}+\sum_{t=1}^n{n \choose t}\cdot 3^t\cdot(-1)^{n-t}\\&=(-1)^n+3\sum_{t=1}^n{n \choose t}\cdot 3^{t-1}\cdot(-1)^{n-t}\\&=3k+(-1)^n\end{array}$$ where $$k=\sum_{t=1}^n{n \choose t}\cdot 3^{t-1}\cdot(-1)^{n-t}$$

miracle173
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  • Note that $(3-1)^n\equiv (-1)^n\mod 3$ and you are done. –  Nov 11 '21 at 09:19
  • @PatBol that was not the question of the OP. The OP asked how to show $(3−1)^ =3+(−1)^$ and it seems that the OP (or the comment he is referencing) wants to avoid modular arithmetic. – miracle173 Nov 11 '21 at 10:52