Lemma 1: Two consecutive odd positve integers must be relatively prime. (straightforward proof which I'll omit)
Lemma 2: x©÷=2^n - 1 with x and n pos. integers has a unique solution. Proof: If n=1 then x=1. If n>1 then 2^n = 0 mod 4 so 2^n - 1 = 3 mod 4 but any perfect square = 0 or 1 mod 4 so no solutions.
Lemma 3: x©÷=2^n + 1 with x and n pos. integers has a unique solution. Proof: x©÷-1=2^n implies (x-1)(x+1)=2^n so (x-1) is a power of 2. let x-1=2^k then 2^k(2^k + 2)= 2^n. So 2^(2k) + 2^(k+1)=2^n which implies that 2^(n-k-1) - 2^(k-1)=1. The only powers of 2 with a difference of 1 are 1 and 0 so k=1, n=3, and x=3 only.
Since m=2 does not work, m must be odd. let m=2k + 1. So n©÷=(2^(2k) - 1)/(2k+1) = (2^k -1)(2^k +1)/(2k+1). Either m divides (2^k - 1) or it divides (2^k + 1).
case 1: If m divides 2^k - 1 then 2^k -1= mw which implies that n©÷=w(2^k + 1). If w is not a perfect square then w and (2^k +1) are not relatively prime and have a common factor v¡Á1. So v is a common factor for (2^k + 1) AND (2^k - 1) which is impossible by lemma 1. If w is a perfect square then so is (2^k + 1) so by lemma 3, k=3 forcing m=7 only.
case 2: If m divides 2^k +1 then 2^k + 1=mw which implies that n©÷=w(2^k - 1). Again if w is not a perfect square then w and (2^k - 1) are not relatively prime and have a common factor v¡Á1. So v is a factor of both (2^k + 1) and (2^k - 1) which by lemma 1 is impossible. If w is a perfect square then so is (2^k - 1) which by lemma 2 forces k=1 and consequenly m=3 only.
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Posted by Dennis
on 2006-09-24 15:40:02 |
a solution
