3

Problem: Given $2018$ elements not necessarily distinct from $\mathbb{Z_4}^9=\mathbb{Z_4} \times \mathbb{Z_4}\times \mathbb{Z_4}\times \mathbb{Z_4}\times \mathbb{Z_4}\times \mathbb{Z_4}\times \mathbb{Z_4}\times \mathbb{Z_4}\times \mathbb{Z_4}$, Prove that there exist 4 elements from these $2018$ which sum to $\text{0}= (0,0,0,0,0,0,0,0,0)$. Obviously you cant use an element more than the number of times it is available.

Origin: This problem originates from a 'number theory' question: Given $2018$ integers all of whose prime factors are $\leq 23$ show that there are $4$ whose product is a fourth power.

My attempt: The only slightly fruitful approach I had was to attach this problem coordinate-wise.. Since, there are 2018. There will be an integer in $\mathbb{Z_4}$ that occurs at least $2018/4$ that is $505$ times. Now starting with these $505$ we go to the next coordinate and so on. But in this way I will need about $4^9 > 2018$ integers to do my job.

I feel a more global approach is needed. I tried some contradiction but that also didnt bear any fruit.

Please help. Thanks.

Subham Jaiswal
  • 3,795
  • 1
    We can take four times the element $(0,0,\ldots ,0)$. – Dietrich Burde Jun 30 '20 at 08:50
  • 1
    @DietrichBurde this element may not be available in the given 2018 elements. – Subham Jaiswal Jun 30 '20 at 09:05
  • 1
    Perhaps there is a stronger statement over $\Bbb Z_2$ that can be used here. For instance, it's clear that (in the original number theoretic version) we can find a pair of elements whose product is a perfect square, but it's not clear that we necessarily find that same perfect square twice. – Ben Grossmann Jun 30 '20 at 09:19

1 Answers1

2

In my opinion this problem is more about rather than .

Anyway, a pigeonhole exercise of this type can often be done in two stages. In the first we produce a lot of squares. Here we would first produce a lot of pairs of vectors belonging to $\{0,2\}^9$.

Observation. Given $513$ vectors from $\Bbb{Z}_4^9$ we can find among them two vectors $u$ and $v$ such that their sum has only even components. In other words $u+v\in\{0,2\}^9$.

Proof. This is basic pigeon-holing. There are $2^9=512$ parity combinations, so we can find $u,v$ such that $u_i\equiv v_i\pmod2$ for all $i$. This pair works.

Then we can proceed with the main task. First run the following algorithm:

  1. Initialize $S$ to be the list of the given $2018$ vectors. Initialize a counter $n=0$, and initialize another list $L$ to be empty.
  2. If more than $512$ elements remain in the list, the observation tells us that we can find a pair of entries $u,v$ from the list $S$ such that $u+v\in\{0,2\}^9$. Otherwise, the algorithm stops here.
  3. Increment $n\leftarrow n+1$, set $u_n=u,v_n=v$. Remove $u$ and $v$ from the list $S$ and add the pair $(u_n,v_n)$ to the list L.
  4. Go to step 2.

After this algorithm has run its course, there are at most $512$ vectors remaining in $S$. Therefore we have removed at least $2018-512=1506$ elements from the list $S$. Thus we can conclude that the list $L$ now has $n=753>512$ pairs $(u_i,v_i),i=1,2,\ldots,n$ such that for all $i$ we have $u_i+v_i\in\{0,2\}^9$.

We are now well places to reapply the observation to the collection of $753$ vectors $(u_i+v_i)/2$. The observation gives us pairs $(u_i,v_i)$, $(u_j,v_j)$, $i\neq j$, such that $(u_i+v_i+u_j+v_j)/2\in\{0,2\}^9$.

Therefore all the components of $u_i+v_i+u_j+v_j$ vanish modulo four.

Strange. Looks like I didn't need nearly all of the $2018$ vectors, $1537$ would seem to suffice to produce $513$ pairs?

Jyrki Lahtonen
  • 140,891