Let $b$ be a positive integer. Determine all $2002$-tuples of non-negative integers $(a_{1}, a_{2}, \cdots, a_{2002})$ satisfying \[\sum^{2002}_{j=1}{a_{j}}^{a_{j}}=2002{b}^{b}.\]
Problem
Source:
Tags: inequalities, Diophantine Equations
matex (L)(L)(L)
16.11.2008 03:04
how to solve this problem? Is there any nice solution?
Peter
16.11.2008 09:50
There should be. I can prove that when $ b\ge 2002$, then $ a_1 = ... = a_{2002} = b$ is the only solution (simply by inequalities), but I have no idea how to handle $ b < 2002$. Yimin Ge, do you know more? You're from Autstria...
Rust
16.11.2008 16:27
Let $ c=max_j a_j$. Then $ c^c\le S=\sum_j a_j^{a_j}\le 2002 c^c$. If $ c\ge 737$, then $ 2002(c-1)^{c-1}<c^c$, therefore $ b\ge c\to a_j=c \ \forall j$. When $ b<737$ may be another solutions. For example $ b=1,a_1=a_2=...a_k=1,a_{k+1}=...=a_{2002}=0.$
FelixD
16.01.2009 14:20
In the competition were this problem appeared, there was the additional condition $ b>800$.
Peter
16.01.2009 15:13
FelixD wrote: In the competition were this problem appeared, there was the additional condition $ b > 800$. That is interesting information! Do you have a solution?