
Re: Reciprocals of integers summing to 1
Posted:
Nov 17, 2012 2:44 PM


On Nov 17, 11:52 am, luir...@yahoo.com wrote: > El viernes, 16 de noviembre de 2012 00:39:24 UTC4:30, CharlieBoo escribió: > > > For each n, what are the solutions in positive integers (or in > > integers) to (1/X1)+(1/X2) + . . . + (1/Xn)=1 ? > > There are infinitely many different solutions. But I am not sure that it is valid for all n. > Take k primes and do the following sum: 1/p1 + 1/p2 + 1/p3 + ...+ 1/pk = S < 1 > Now make Q = 1  S. > By the theorem of the Egyptian fractions, Q always can be decomposed as: > Q = 1/x1 + 1/x2 + 1/x3 +....+ 1/xj. > I am not sure that it is possible, ever, that k+j = n. > Ludovicus
There are an infinite number of sets of positive integers whose reciprocals sum to 1:
N = # of terms. Term # M where M = 1 to N = ( If M < N then (M+1)! / M Else M ! )
CB

