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 2005-04-13, 03:00 #5 wblipp     "William" May 2003 New Haven 26×37 Posts One small solution is p=28, q=73, r=4. It's easy to generate an infinite number of solutions from Mersenne factors, although they grow rather quickly. When Robert Silverman left off, we have a solution for any choices of m. k, q, and h such that m=k+2h and (kq+1)=2m-h If we let z=m-h, we have a solution for any values of z, k, q, and h with k=z-2h+h and kq=2z-1 Eliminating k, we see that we have a solution for any choice of z, q, and h with q=(2z-1)/(z-2h+h) z cannot be a prime number because the denominator is smaller than z and all factors of 2p-1 are of the form 2ap+1, and therefor larger than p. So let z=xy and pick the denominator to be a factor of 2x-1. So to generate a solution: 1. Pick a value for h. 2. Pick x to be a factor of 2h-h+1 3. Pick the denominator to be a factor of 2x-1 4. Solve for y from the denominator choice 5. z=x*y 6. Work backwards to p, q, and r. For example 1. h=2 2. 2h-h+1=3, so pick x=3 3. 2x-1=7, so pick the denominator=7 4. y = 3 5. z = xy = 9 6. q = (2z-1)/denominator = 73 7. m = z+h = 11 8. k = denominator = 7 9. r = 2h= 4 10. p = kr = 28