1/x + 1/y + 1/z = 1/n
Tim
1/x + 1/y + 1/z = 1/n and 1<=x<=y<=z ,
what is the maximal possible value of z ?
Thanks
Tapio Hurme
"Tim" <tim_b...@my-deja.com> wrote in message
news:93880d6b.03062...@posting.google.com...
Consider Pythagorean quartets - for example:
a^2+b^2+c^2=d^2 Divide both sides by (abcd)^2
As there exist infinite many Pythagorean quartets, what you can say about
the maximal possible value of z, which can be because of the symmetry :
(bcd)^2 or (acd)^2 or (abd)^2 and - of course - n=(abc)^2 >=1
Tapio
Erick Bryce Wong
It's natural to conjecture that z <= n(n+1)(n^2+n+1) based on
1/(n+1) + 1/(n^2+n+1) + 1/n(n+1)(n^2+n+1) = 1/n, and I think
this is in fact optimal. Here's a proposed proof:
Let a/b = 1/n - 1/x = (x-n)/nx. I claim that z <= (b^2 + b)/a.
To see this, rewrite 1/y + 1/z = a/b as (ay-b)(az-b) = b^2 > 0.
Since all quantities are positive integers, we have az-b <= b^2
from which the bound follows.
Now we know n < x <= 3n (in fact if x = 3n we only get the trivial
solution x = y = z = 3n). Taking a = x-n, b = nx (note that a/b is
not necessarily in lowest terms), the upper bound is nx(nx+1)/(x-n).
Rewriting in terms of a = x-n >= 1, the upper bound becomes:
z <= 2 n^3 + n + n^2 ( a + (n^2+1)/a ).
The right-hand side, as a function of a, has a unique local minimum
at a = sqrt(n^2+1), and thus achieves its maximum value at one of
the endpoints, a=1 or a=2n. Since n^2+1 >= 2n, it's easy to see
that a=1 is the higher of the two, which finally gives us:
z <= 2 n^3 + n + n^2 (n^2 + 2) = n(n+1)(n^2+n+1).
Does the result generalise to longer sums of Egyptian fractions?
-- Erick
Dan
Fixing n=1 , trying to generalise leads to the sequence
a(1)=1 , a(n)=a(n-1)^2+a(n-1) and this is sequence A007018 in the OEIS
:
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A007018
See also http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000058
that mentions egyptian fractions .
Dan