Dr. Katz DEq Homework Solutions 93

Dr. Katz DEq Homework Solutions 93 - either (6.6.0.15) l...

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Algebraic Solutions of Differential Equations 93 If a~Z, a<0, then --1 Rx,(_a),,. 1 for all p sufficiently large. Then for P any A invertible mod N, the limit formula gives (6.6.0.12) 1 > (bA) > (cA) >0. Just as above, replacing A by -A shows (bA)= (cA) for all invertible A modulo N. Hence c-b lies in Z, hence (as a~Z), c-b-a lies in Z, another contradiction, so (6.6.0.7) does not occur. Thus we have shown that ar bq~Z, and that for every A invertible mod N, we have either (6.6.0.13) l >(b A) > (cA) > (aA) >O or (6.6.0.14) 1 > (aA) > (cA) >= (b A) > O. We next show that c-aq~Z and c-bq~Z. As by hypothesis a-br we cannot have both c- a ~ Z and c- b ~ Z. Suppose that c- a ~ Z. Then c-br hence for all A invertible mod N, A c-A br
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Unformatted text preview: either (6.6.0.15) l &gt; (b A) &gt; (cA)=(aA) &gt;O Or&quot; (6.6.0.16) 1 &gt; (aA) = (cA) &gt; (bA) &gt;0. But (6.6.0.15) cannot hold for both A and -A. Hence there exists a A invertible modulo N for which (6.6.0.16) holds. Rewriting (6.6.0.1) for this A via the limit formula (6.5.3.1), we get (6.6.0.17) 1&gt; lim 1Rp(-a)= lim LRp(--c)&gt; lim 1---Rp(-b)&gt;0. p~ p p~ p p~ p pd -~ (N) pA =- 1 (N) pA =. 1 (N) Thus for all p sufficiently large with pA = 1 (N), we have (6.6.0.18) 1 Rp(_C)&gt;L Rp(_b)&gt;O&quot; P P This is incompatible with the first of the two following inequalities, one or the other of which holds for any sufficiently large prime in virtue of (6.4.0) (6.6.0.19) l~ Rv(-b)&gt;lRp(-c)&gt;lRp(-a), P P P (6.6.0.20) 1Rp(-a)&gt; l~ Rp(-C)&gt;__lRp(-b). P P P...
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This note was uploaded on 12/21/2011 for the course MAP 4341 taught by Professor Normankatz during the Fall '11 term at UNF.

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