Existence of solutions to (n,n+1,n+1)
THEOREM (Jaroslaw Wroblewski, July 7, 2002):
For any n there exists a solution to (n,n+1,n+1). However it may
have some (but not all!!!) of left terms equal to some of right terms,
so in fact it may reduce to a solution to (n,k,k) for a k<=n+1.
PROOF:
Let N be a large positive integer to be specified later.
Consider all sytems of positive integers
(a1,a2,...,an+1)
such that
N>=a1>=a2>=...>=an+1.
There are at least
s = Nn+1 / (n+1)!
such systems, as there are Nn+1
(n+1)-tuplets of positive
integers not exceeding N, and any nonincreasing sequence corresponds
to at most (n+1)! such tuplets.
For each of the systems, consider the sum
a1n + a2n + ... + an+1n
being a number not exceeding
S = (n+1) * Nn .
For large N (namely N>(n+1)*(n+1)!) we have
s > S
and there are more considered systems than possible sums associated
with them. By pigeon-hole principle there exist two different systems
having the same sum associated with them, giving
a1n + a2n +
... + an+1n =
b1n + b2n +
... + bn+1n
with
N>=a1>=a2>=...>=an+1 and
N>=b1>=b2>=...>=bn+1.
Although systems
(a1,a2,...,an+1) and
(b1,b2,...,bn+1) are
different, they may have some common elements.
This ends the proof of the theorem.
REMARKS:
Proof is not constructive at all. It proves there exists a solution to (n,n+1,n+1), but gives no indication how to find a particular example.
As a corollary we get the following (the first place where the estimates in power table are proven to be improvable):
There exists a solution to (24,25,25).
It may have some (but not all) common terms on left and right sides. Such solution involves numbers not exceeding
25*25!+1=387780251083274649600000001 (27 digits)
Jaroslaw Wroblewski, July 7, 2002
EXTENSION:
We can extend Jaroslaw Wroblewski's proof to show that there exist
linked solutions to
a1,1n + a1,2n + ... + a1,n+1n
= a2,1n + a2,2n + ... + a2,n+1n
= ... = am,1n + am,2n
+ ... + am,n+1n
for arbitrarily large m.
The number of systems s = Nn+1 / (n+1)! and the number
of sums S = (n+1) * Nn are as before.
If the number of systems is more than m-1 times the number of sums,
then by the pigeon-hole principle there is at least one sum
that must have m systems associated with it. This occurs when s>(m-1)*S
or N>(m-1)*(n+1)*(n+1)!
Stuart Gascoigne, August 5, 2002
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