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Let "a" be an odd integer. Prove that a^{2n}(is congruent to) 1 (mod 2^{n+2})

Attempt: By using induction:

Base Case of 1 worked.

IH: Assume a^{2k}(is congruent to) 1 (mod 2^{k+2})

this can also be written: a^{2k}= 1 + (l) (2^{k+2}) for some "l"

IS: a^{2k+1}= a^{2k°2}= (a^{2k)2}

Now I took 1 + (l) (2^{k+2}) and substituted it into (a^{2k)2}and expanded:

1 + l ( 2^{k+3}+ (l) 2^{2k+4}) is what I obtained after expanding and then simplifying it. But I know this isn't what I have to obtain when I try and show the K+1 case. What am I missing?

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# Test Today Quick Number Theory Question

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