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Prove that:

$$a^{2^n} \equiv 1\pmod{2^{n+2}}$$

I have done the base case, and then have an assumption that:

$$a^{2^k} \equiv 1 \pmod {2^{k+2}}$$

I wish to prove the last step, that:

$$a^{2^{k+1}} \equiv 1 \pmod{2^{k+3}}$$

How would I do this?

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First of all, you should specify that $a$ is odd, and $n > 0$.

And then, use the definition of the congruence relation $$ a^{2^k} \equiv 1 \pmod{2^{k+2}} $$ to write $$ a^{2^{k}} = 1 + c 2^{k+2}, $$ for some $c$, then take the square.

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I'm assuming the you forgot to specify that $a$ is odd. From the induction hypothesis you know that $a^{2^{k}}=t 2^{k+2} + 1$ for some integer $t$. From this we get

$$a^{2^{k+1}}=(a^{2^{k}})^{2}=(t 2^{k+2} + 1)^2=t^2 2^{2k+4} + t2^{k+3} + 1$$

But $2^{2k+4} \equiv 0 \mod(2^{k+3})$ and so $$t^2 2^{2k+4} + t2^{k+3} + 1 \equiv 1 \mod(2^{k+3})$$

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Lemma $\rm\ k>0,\,\ 2^k\!\mid b-1\:\Rightarrow\: 2^{k+1}\!\mid b^2-1$

$\begin{eqnarray}\rm{\bf Proof}\quad k>0,\,\ \color{#0A0}{2^k}\!&\mid&\rm\color{#0A0}{b-1}\:\Rightarrow\: \color{#C00}2\cdot \color{#0A0}{2^{k}}\!\mid \color{#0A0}{(b-1)}(\color{#C00}{b+1})\,=\,b^2\!-1\\ \\ \rm since\ \ \ \ 2&\mid&\rm b-1\:\Rightarrow\:\color{#C00}2\mid\color{#C00}{b+1} =(b-1) + 2\end{eqnarray}$

Remark $\ $ Yours is the special case $\rm\, b = a^{2^n},\,\ k = n+2.$

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To the downvoter: I can explain further if something above is not clear. This viewpoint shows clearly the source of the extra factor of $2$ at each step. – Math Gems Feb 12 '13 at 22:33

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