# 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