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Egyptian Fractions

  1. Aug 10, 2007 #1
    The following is a well-known unsolved problem :

    If n is an integer larger than 1, must there be integers x, y, and z, such that 4/n=1/x+1/y+1/z?
    A number of the form 1/x where x is an integer is called an Egyptian fraction.
    Thus, we want to know if 4/n is always the sum of three Egyptian fractions, for n>1.
    .....................................................................................................
    So, if 4/n=1/x + 1/y + 1/z, then there exists a third degree polynomial x^3 + ax^2 + bx + c where 4/n=b/c for integers b and c. This is because 1/x + 1/y + 1/z=(z(x+y) +xy)/xyz. And the polynomial (a+x)(a+y)(a+z)=a^3 + (x+y+z)a^2 + (z(x+y)+xy)a +xyz, which proves the statement when b=z(x+y) +xy and c=xyz.

    Your thoughts?
     
  2. jcsd
  3. Aug 11, 2007 #2
    What about 1/2 +1/5 +1/7 = 59/70, how does that work out?

    You are confused, as I read you, about The Erdős–Straus conjecture:
    The form 4/N can always be expanded into three Egyptian fractions.

    However, there has got to be restrictions on N, which Wikipedia puts at [tex] N \geq 2[/tex] and others just ignore. However, 2 will not work: 2 = 1/x + 1/y +1/z, if we are speaking of positive all different intergers for x,y,z. Thus the first acceptable case is 4/3 = 1+1/4+1/12, and then 4/4 = 1/+1/3+1/6; and 4/5 = 1/2+1/5+1/10; 4/6 =2/3=1/3+1/4+1/12; 4/7 =1/2+1/15 + 1/210. Maybe Erdos, etc, were assuming 4/N was less than 1 anyway.
     
    Last edited: Aug 12, 2007
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