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Prove for all natural numbers $n$ such that $2^{3^n}+1$ is not divisible by $3^{n+2}$.

My working for not divisible: Induction proof

Base case: n = 0 $$3^{0+2}=9$$ and $$2^{3^0}+1=3$$ 9 cannot divide 3, so base case is true.

Assume n=k is true. That means $B=3^{k+2}$ does not divide $A=2^{3^k}+1$. Now prove $3^{k+3}$ does not divide $2^{3^{k+1}}+1$.

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$: $$(2^{3^k})^3+1=(2^{3^k}+1)((2^{3^k})^2-2^{3^k}+1)$$ $$=A((2^{3^k})^2-A+2)$$

$$3^{k+3}=3\times 3^{k+2}$$ $$=3B$$

That's where I need help!

I want to show that $A((2^{3^k})^2-A+2)$ is not divisible by $3B$ by using hypothesis "$A$ is not divisible by $B$".

James Harden
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    You've made an error. $(2^{3^k})^2 -2^{3^k} + 1 \ne (2^{3^k})^2 - (2^{3^k} + 1)$. – eyeballfrog May 12 '21 at 13:57
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    Related: https://math.stackexchange.com/questions/1012410/the-number-23n1-is-divisible-by-3n1-and-not-divisible-by-3n2 ;https://math.stackexchange.com/questions/4135192/3n2-does-not-divide-23n1 – Etemon May 12 '21 at 14:12
  • We didn't learn about Lifting The Exponent (LTE) in class, so that's why I can't that for my answer. That's why I can't find any good proof without LTE – James Harden May 12 '21 at 14:15

1 Answers1

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Hint: It is easy to see that $$3~|~(2^{3^k})^2-2^{3^k}+1$$ but $$9\nmid(2^{3^k})^2-2^{3^k}+1.$$

Note: To complete the proof, use the Unique Factorization Theorem of integers and count the multiplicity of the prime factor $3$. (This theorem requires only the Division Algorithm to prove.)

An alternative formal way to show that $3B\nmid A\cdot((2^{3^k})^2-A+2)$ is to use the 3-adic order: Compare the $3$-adic order $\nu_3$ of the two integers, one has $$\nu_3(3B)=\nu_3(3)+\nu_3(B)=1+\nu_3(B)$$ and $$\nu_3(A\cdot ((2^{3^k})^2-A+2))=\nu_3(A)+\nu_3((2^{3^k})^2-A+2)=\nu_3(A)+1,$$ where the latter equality follows from the above suggestions and comments.

Now the induction hypothesis $B\nmid A$ shows that $$\nu_3(B)>\nu_3(A)$$ $$\Rightarrow \nu_3(B)+1>\nu_3(A)+1$$ $$\Rightarrow 3B\nmid A\cdot((2^{3^k})^2-A+2),$$ completing the case $n=k+1$ and the proof by induction.

Pythagoras
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