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trap101
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Test Today...Quick Number Theory Question
Let "a" be an odd integer. Prove that a2n (is congruent to) 1 (mod 2n+2)
Attempt: By using induction:
Base Case of 1 worked.
IH: Assume a2k (is congruent to) 1 (mod 2k+2)
this can also be written: a2k = 1 + (l) (2k+2) for some "l"
IS: a2k+1 = a2k°2 = (a2k)2
Now I took 1 + (l) (2k+2) and substituted it into (a2k)2 and expanded:
1 + l ( 2k+3+ (l) 22k+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?
Let "a" be an odd integer. Prove that a2n (is congruent to) 1 (mod 2n+2)
Attempt: By using induction:
Base Case of 1 worked.
IH: Assume a2k (is congruent to) 1 (mod 2k+2)
this can also be written: a2k = 1 + (l) (2k+2) for some "l"
IS: a2k+1 = a2k°2 = (a2k)2
Now I took 1 + (l) (2k+2) and substituted it into (a2k)2 and expanded:
1 + l ( 2k+3+ (l) 22k+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?