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I know this question has been asked before but I have a version of an induction proof which I dont know if its correct or not. Moreover, I am interested in whether my proof-writing is okay or if you have any suggestions on how I can improve it.

Proof. $(i)$ For the base case put $n=1$, which yields $$ 3^{3\cdot 1+3}-26\cdot 1-27=\cdots =676,$$ which is divisible by $169$, since $676=4\cdot 169$.

$(ii)$ For the induction step we now assume that the statement is true for $n=m$. We need to show that the statement is true for $n=m+1$. That is, we need to show that $$169\vert 3^{3(m+1)+3}-26(m+1)-27.$$

Consider $$ 3^{3(m+1)+3}-26(m+1)-27.$$ This expression may be written as $$3^{3(m+1)+3}-26(m+1)-27=3^{3m+6}-26m-26\\-27=(3^{3m+3}-26m-27)+(3^{3m+6}-3^{3m+3}-26).$$ By the induction step we know that the first term $$(3^{3m+3}-26m-27)=169k$$ for some $k\in \mathbb{N}$. Now we only need to prove that the second term $$R(m)=3^{3m+6}-3^{3m+3}-26$$ is divisible by $169$.

Note that $169=13\cdot 13$ and $$R(m)=3^{3m+6}-3^{3m+3}-26=3^{3m+3+3}-3^{3m+3}-26=3^{3m+3}\left(3^{3}-1 \right)-26=\\2\cdot 13\cdot \left( 3^{3m+3}-1 \right).$$ If we divide $R(m)$ by $169$ then $$\dfrac{2\cdot 13\cdot \left( 3^{3m+3}-1 \right)}{169}=\dfrac{2\cdot \left( 3^{3m+3}-1 \right)}{13}.$$ Thus $$169\vert R(m)$$ if and only if $$13\vert 3^{3m+3}-1$$. $$13\vert 3^{3m+3}-1\Leftrightarrow 3^{3m+3}\equiv 1~ (mod~13) \Leftrightarrow 27^{m+1}\equiv 27\cdot 27^{m}\equiv 1\cdot 27^{m}\equiv 27^{m}\equiv 1~(mod~13).$$ Byt the binomial theorem we get that $$27^{m}=(26+1)^{m}=\sum_{k=0}^m {m \choose k}26^{k}\cdot 1^{m-k}=\sum_{k=0}^m {m \choose k}26^{k}=1+\sum_{k=1}^m {m \choose k}(13\cdot 2)^{k} $$ where $13$ divides each term ${m\choose k}(13\cdot 2)^{k}$ for all $k\ge 1$. Thus $$1+\sum_{k=1}^m {m \choose k}(13\cdot 2)^{k}=1+13\cdot c$$ for some $c\in \mathbb{N}$. That is $27^{m}\equiv 1~(mod~13)$ for all $m\in \mathbb{N}$.

Thus, we have shown that $$169\vert 3^{3(m+1)+3}-26(m+1)-27$$ and we have therefore proven (by induction) that for positive integers $n$ we have that $$169\vert 3^{3n+3}-26n-27.$$

Victor Galeano
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  • Your binoial theorem stuff is redundant after showing that $3^{3m+3}\equiv_{13}1$ (which can also be shortened to just $3^{3m+3} = 27^{m+1}\equiv_{13} 1^{m+1} = 1$). They show the same thing, and you only need one of them. Apart from that, after a quick look through, this seems like a decent proof. – Arthur Nov 28 '19 at 13:07
  • Thats true! Now I feel a bit stupid. Hehe. I used that $27\equiv 1~(mod~13)$ and basic modular arithmetic gives that $27^{m+1}\equiv 1^{m+1}\equiv 1~(mod~13)$. Well, redundant as it is it was very informative to me. Hehe. Thanks! :) – Victor Galeano Nov 28 '19 at 14:16
  • Dupe of this – Bill Dubuque Aug 23 '23 at 15:43

1 Answers1

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Its better to eliminate the exponent of $3$

If $f(n)=3^{3n+3}-26n-27$

$$f(n+1)-3^3f(n)=-26(n+1)-27+3^3(26n+27)=27(3^3-1)-26+26n(3^3-1)$$ which is clearly divisible by $13^2$

If $169|f(n)\implies169|f(n+1)$

Alternatively using Binomial Theorem, $$3^{3n+3}=(1+26)^{n+1}=1+26(n+1)+\sum_{r=2}^{n+1}\binom{n+1}r26^r\equiv1+26n+26\pmod{26^2}$$

J. W. Tanner
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lab bhattacharjee
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