I would like to prove that
$$a^{2^n}\equiv 1 \pmod {2^{n+2}}$$
I tried induction but could not get it. Thank you very much!
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6You need some hypothesis on $a$, e.g., it's false for $a=2$. – Gerry Myerson Feb 25 '13 at 00:05
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$$0^{2^n}\neq 1\pmod{2^{n+2}}\ldots$$ – DonAntonio Feb 25 '13 at 00:06
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1isn't this false? take $n=1, \quad a=2$. Then $4 = {1} \mod {8}$ which is false? Or am I missing something – Ben Feb 25 '13 at 00:07
2 Answers
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$$a^{2^n}-1=(a-1)(a+1)(a^{2}+1)\cdots(a^{2^{n-1}}+1)$$
If $a$ is odd then each term in the factorisation is even (i.e. divisible by $2$), there are $n+1$ such terms hence $a^{2^n}-1$ is divisible by $2^{n+1}$. Since either $a-1$ or $a+1$ is a multiple of $4$, we have an additional factor of $2$, therefore $a^{2^n}-1$ is divisible by $2^{n+2}$
L. F.
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Assume $a$ is odd! And note we have to start at $n=1$. The base step is the familiar fact that the square of an odd number is congruent to $1$ modulo $8$. We now deal with the induction step.
Suppose that we know the result for $k$. So $a^{2^k}=1+b2^{k+2}$. Square both sides. On the left, we get $a^{2^{k+1}}$. On the right, we get the extra factor of $2$ from the middle term in the expansion $(s+t)^2=s^2+2st+t^2$.
André Nicolas
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