# im 2 ix and positive or negative integer coe cients

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Unformatted text preview: 2k, 2k + 1, 2k + 2, and 3k, 3k + 1, 3k + 2, 3k + 3, or in general jk, jk + 1, . . . jk + j . For j k 1 these blocks overlap and no integers are left out. In the last example 24, 25 2 I0 implies 48, 49, 50 2 I0 which implies 72, 73, 74, 75 2 I0 and 96, 97, 98, 99, 100 2 I0 , so we know the result holds for n0 = 96. In fact it actually holds for n0 = 34 but it is not important to get a precise bound. To show that there are two consecutive integers, we cheat and use a fact from number theory: if the greatest common divisor of a set Ix is 1 then there are integers i1 , . . . im 2 Ix and (positive or negative) integer coe cients ci so that c1 i1 + · · · + cm im = 1. Let ai = c+ and bi = ( ci )+ . In words the ai i are the positive coe cients and the bi are 1 times the negative coe cients. Rearranging the last equation gives a1 i1 + · · · + am im = (b1 i1 + · · · + bm im ) + 1 and using Lemma 1.15 we have found our two consecutive integers in Ix . 25 1.5. LIMIT BEHAVIOR While periodicity is a the...
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