If $a$ is odd then $a^{2n} \equiv 1 \pmod{2^{n+2}}$ for all $n≥1$
I try using induction but I'm not get it.Thank you
If $a$ is odd then $a^{2n}\equiv 1\pmod{2^{n+2}}$ for all $n≥1$
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number-theory
elementary-number-theory
modular-arithmetic
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0Perhaps you mean $a^{2^n}$ ? – 2017-02-17
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0I'm not sure. If it is $a^{2^n}$ , It is true? – 2017-02-17
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0@TheetaNonatee Yes. – 2017-02-17
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0See http://math.stackexchange.com/questions/2147520/show-that-52e-2mod2e-1-mod2e – 2017-02-17
1 Answers
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This is false. Take $a=3, n=3$. Note that $$3^{6}=729 \equiv 25 \not \equiv 1 \pmod {2^{5}}$$ And $2^{5}=32$. Thus your claim is false.
Perhaps you mean $$a^{2^{n}}$$ in which case it follows from Euler's Formula and the fact that $$x^2+1 \not \equiv 0 \pmod {8}$$
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0Right. You beat me to it in correcting what the statement should have been. – 2017-02-17