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I've been stuck on this mathematical proof for quite some time now.

So far I have tried using the division algorithm to prove the theorem by letting

$n = 3 \times q + r\ |\ q \epsilon N$

This gives me three different cases for values of n.

(i) $n = 3 \times q + 0$

(ii) $n = 3 \times q + 1$

(iii) $n = 3 \times q + 2$

Basically what I am thinking now is that if I can prove that all three of these cases are divisible by 3 then $(14^{2n}-1)$ must be divisible by all n = 0, 1, 2, 3, ...

So I try to prove the theorem for case (i)

$14^{3\times q \times 2}-1 = 14^{6 \times q} - 1$

Now I'm stuck. What steps can I take to prove that this is divisible by 3?

Thanks in advance for any help.

  • 1
    Are you familiar with modular arithmetic? – 2017-02-18
  • 1
    Since $14\equiv -1 \pmod{3}$, then $14^{2n}\equiv 1\pmod{3}$. – 2017-02-18

4 Answers 4

0

If a square $a^2$ is not a multpiple of $3$, then $a^2-1$ is.

Indeed, $a^2-1=(a+1)(a-1)$. Exactly one of the numbers $a-1$, $a$, $a+1$ is a multiple of $3$ and the result follows.

1

For $n=0$ we have $14^{2n}-1=0$, and $13$ divides $0$. For $n\ge 1$ \begin{align*} 14^{2n}-1&=\left( 14^2\right)^n-1\\ &=(14^2-1)\left[(14^2)^{n-1}+(14^2)^{n-2}+\ldots+14^2+1\right]\\ &=(14-1)(14+1)\left[(14^2)^{n-1}+(14^2)^{n-2}+\ldots+14^2+1\right]\\ &=13(14+1)\left[(14^2)^{n-1}+(14^2)^{n-2}+\ldots+14^2+1\right] \end{align*} So $13$ divides $14^{2n}-1$.

0

$14=2$ mod 3 implies $14^2=14^{2n}=1$ mod 3. We deduce that $14^{2n}-1=0$ mod 3.

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$14\equiv-1\mod 3$ and $14^{\color{red}2n}\equiv\bigl( (-1)^2\bigr)^n=1$.