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Home » High School Contests & Programs » Other Problem Solving Topics » Problems From the Book » Valuations and the order of an element
 
14.4
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Altheman
Birch & Swinnerton Dyer
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#1
14.4
Bulgaria 1997
Find all positive integers m,n for which n|m^{2\cdot 3^n}+m^{3^n}+1.
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PostPosted: Wed Jul 22, 2009 6:32 pm  Back to top 
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Poincare Conjecture
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First consider the case where n is not a power of 3. Take some prime p|n, p\neq3. Then if m^{3^n}\equiv1 (mod p), m^{2*3^n}+m^{3^n}+1\equiv3 (mod p). But p does not divide 3, so m^{3^n} is not congruent to 1 (mod p). But n|m^{2*3^n}+m^{3^n}+1|m^{3^{n+1}}-1, so m^{3^{n+1}}\equiv1 (mod p). Thus ord_p(m) divides 3^{n+1} and does not divide 3^n, so it is 3^{n+1}. Thus 3^{n+1}|p-1, so p>3^{n+1}>n. But p|n, a contradiction. Thus n is a power of three. However, x^2+x+1 does not have a root mod 9, so n=1 or n=3. If n=1, then m can be anything. If n=3, 3|m^{54}+m^{27}+1. 3 does not divide m, so this is equivalent to 3|m^{81}-1. m^2=1 (mod 3), so this is equivalent to 3|m-1, or m\equiv1 (mod 3). Thus the solutions are n=1, or n=3, m\equiv1 (mod 3).
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