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Using Weak Mathematical Induction, I have to show that, for all integers $n \geq 1$, $8|3^{2n} -1$

I really don't know how to go about solving this problem. Currently I only have the base case and the Inductive Hypothesis:

Base Case: For $n = 1$ $$8|3^{2(1)}-1 = 8|8 = 0$$

Inductive Hypothesis: Assume true for $n = k$ $$8|3^{2k}-1$$

Inductive Step: I want to show that the statement is true for $n = k+1$ so $$8|3^{2(k+1)}-1$$

This is where I am currently stuck. Any help would be really helpful. Thanks

Martin Sleziak
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  • I am not sure if I have the right thought process for this, but in general for a ≡ b (mod n) can be written as n|(a-b) similarly for this problem I did 3^(2(k+1)) ≡ 1 (mod 8) Then I get stuck again. – Guest Nov 12 '14 at 02:55
  • This thought process deserves to be part of the main question. It is a significant idea that you contributed on your own. Even if it were useless (and it is not!) it would guide people to make more helpful responses. – David K Nov 12 '14 at 04:06
  • You have written some things that are not grammatical, and so show only a partial understanding of the concepts. When you write “$a|b$”, you have made a statement that’s either true or false. It doesn’t have a value. So, “$8|8$” is true, but nothing you have written is equal to zero. – Lubin Nov 12 '14 at 04:40

2 Answers2

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$8$ divides $3^{2k}-1$ then $3^{2k}=8m+1$ for some $m$.

$3^{2(k+1)}-1 = 3^{2k} \cdot 3^2-1 = (8m+1) \cdot 3^2-1 = 8m \cdot 3^2+3^2-1 = 8m \cdot 3^2+8$ which is divisible by 8.

Hugh Denoncourt
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You are on a good path when you consider that $a \equiv b \pmod{n}$ is equivalent to $n | (a-b).$

In the inductive step, you already have $8|(3^{2k}-1),$ so you can write $$3^{2k} \equiv 1 \pmod 8.$$ Now consider how to replace the "?" in $$\frac{3^{2(k+1)}}{3^{2k}} \equiv \; ? \pmod 8$$ and see if these two facts help you show that $$3^{2(k+1)} \equiv 1 \pmod 8.$$

David K
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