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Topic: Arithmetic progression in Z - {m^2+n^5} (Read 2429 times) |
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Aryabhatta
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Arithmetic progression in Z - {m^2+n^5}
« on: Sep 21st, 2008, 11:42pm » |
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Let S be the set of integers which cannot be written in the form m2+n5 for integers m, n.
Show that S contains an infinitely long arithmetic progression.
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teekyman
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Re: Arithmetic progression in Z - {m^2+n^5}
« Reply #1 on: Sep 21st, 2008, 11:59pm » |
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the numbers 0 - 10 squared mod 11 are 0,1,4,9,5,3,3,5,9,4,1 by calculation.
all possible values of n^5 mod 11 are 0 (if n is congruent to 0) or 1/-1. This is true because (n^5)^2 is congruent to 1 by fermats little theorem, and I remember proving that the only quadratic roots of 1 in a prime ring are 1 and -1. (not a hard proof)
Numbers congruent to 7 mod 11 cannot be formed by adding a number congruent to one of (0,1,3,4,5,9) and a number from (0,1,-1), thus the numbers congruent to 7 mod 11 are an example of such an infinite arithmetic progression.
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| « Last Edit: Sep 22nd, 2008, 12:03am by teekyman » |
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Aryabhatta
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Re: Arithmetic progression in Z - {m^2+n^5}
« Reply #2 on: Sep 22nd, 2008, 12:07am » |
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That was quick! Well done.
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