Pigeonhole Principle and Sequences

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Hi guys.

Homework Statement



Given any two sequences of integers of length n, \left{\{a_k\right\}_{k=1}^n,\left{\{b_k\right\}_{k=1}^n,, where 1\le a_k,b_k\le n for all 1\le k\le n, show that there are subsequences of a_k,b_k such that the sum of the elements in each subsequence is equal, i.e.:

If a = 1,6,...,3,2,3,... and b = 3,5,...,2,.., then the subsequence {3,2,3} and {3,5} of a and b would fit the requirements.

Homework Equations



N/A

The Attempt at a Solution



I am completely stuck. So far the best I have is that there are \frac{1}{2}n(n+1) many consecutive subsequences of a, and the same number of b. This means that there are n^2+n many total subsequences of a,b total. The sum of a given consecutive subsequence ranges from 1 (in the case where we select one element as the subsequence, with value 1) all the way to n^2 (in the case where the entire n-length sequence has n's for each of its elements). Therefore, we have n^2+n sums going into n^2 many possible values, implying that at least two sequences have the same value when summed.

The problem is showing that the two sequences (whose existence I just showed) come from a and b separately, and are not both subsequences of just a or just b.

Another thing I tried was consider A_I = a_1+a_2+\cdots+a_I, where 1\le I\le n, with B_I defined similarly. If we assume that no two consecutive sub-sequences from a,b respectively have the same sum, it means that A^*=\left{A_I:1\le I\le n\right\} is of size n, and is disjoint from a similarly defined B^*=\left{B_I:1\le I\le n\right\}. If we can show that there must exist distinct x,y\in A^* and u,v\in B^* such that |x-y|=|u-v| no matter how we choose sequences a,b we are done, but I can't figure out how to do this. It may be a dead end.


Any help would be much appreciated, including hints!

Thanks folks :blushing:
 
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Do the subsequences have to be consecutive (the problem doesn't seem to state this)?
 
Yes, they do.
 
i'm stuck on it too~
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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