A quotient and square problem (Posted on 2006-09-23)

	Submitted by K Sengupta
Rating: 3.3333 (3 votes)
Solution:	(Hide)
( We denote < sign, > sign, < = sign and > = sign respectively by lt, gt, lteq and gteq)) If m is a prime which is not equal to 2, then (2^m -1 ) = (2^(m-1)/2 - 1 )(2^(m-1)/2 + 1 ) is divisible by m in terms of the Fermat's theorem. So, either (2^(m-1)/2 - 1 )or (2^(m-1)/2 + 1 ) is divisible by m. They are relatively prime, since these are successive odd integers. Suppose that m is a odd prime such that and (2^(m-1)/2 - 1 )/m is a square. First, let us suppose that (2^(m-1)/2 - 1 ) is divisible by m. We know that, if x and y are relatively prime positive integers such that their product xy is a perfect square, then x and y separately must be perfect squares. Since, (2^m -1 )/m = ((2^(m-1)/2 - 1 )/p)(2^(m-1)/2 + 1 ); we observe that 2^(m-1)/2 + 1 = a^2, for some integer a gt 2. Thus; 2^(m-1)/2 = a^2 - 1 = (a-1)(a+1) If a-1 = 2^r, a+1 = 2^s for 0 lt r lt s, then: 2^s - 2^r = 2, giving 2^(s-1) - 2^(r-1) = 1 Since, 0 lteq (r-1) lt (s-1), this is possible only if r-1=0, s-1 = 1. Hence, a-1 =2, so that a=3; 2^(m-1)/2 = a^2-1 = 8; (m-1)/2 =3, giving m = 7 Next, let us suppose that 2^(m-1)/2 + 1 is divisible by m. Since, (2^m -1 )/m = (2^(m-1)/2 - 1 )(2^(m-1)/2 + 1 ); we observe, exactly as before that 2^(m-1)/2 - 1 =b^2, for some integer b gt 0. If, (m-1)/2 gteq 2, then this implies that b^2 = -1 (Mod 4), which is not feasible. Hence, (m-1)/2 lteq 1, so that m = 3, since m must be odd. Finally, if m =7, then : (2^(m -1)/2 - 1 )/m = 9 , And if m =3, then : (2^(m -1)/2 - 1 )/m = 1. THUS, THE DESIRED PRIMES ARE 3 and 7.

	Subject	Author	Date
	typos re a solution	Dennis	2006-09-24 18:08:56
	a solution	Dennis	2006-09-24 15:40:02
	fermats little theorem	siddhesh	2006-09-23 19:50:45
	re: prelimanary findings	Dej Mar	2006-09-23 16:54:28
	prelimanary findings	Charlie	2006-09-23 16:05:01