Subfield: Algebra


Question: In an infinite array with two rows, the numbers in the top row are denoted $\ldots, A_{-2}, A_{-1}, A_{0}, A_{1}, A_{2}, \ldots$ and the numbers in the bottom row are denoted $\ldots, B_{-2}, B_{-1}, B_{0}, B_{1}, B_{2}, \ldots$ For each integer $k$, the entry $A_{k}$ is directly above the entry $B_{k}$ in the array, as shown:

| $\ldots$ | $A_{-2}$ | $A_{-1}$ | $A_{0}$ | $A_{1}$ | $A_{2}$ | $\ldots$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $\ldots$ | $B_{-2}$ | $B_{-1}$ | $B_{0}$ | $B_{1}$ | $B_{2}$ | $\ldots$ |

For each integer $k, A_{k}$ is the average of the entry to its left, the entry to its right, and the entry below it; similarly, each entry $B_{k}$ is the average of the entry to its left, the entry to its right, and the entry above it.
Consider the following two statements about a third such array:

(P) If each entry is a positive integer, then all of the entries in the array are equal.

(Q) If each entry is a positive real number, then all of the entries in the array are equal.

Prove statement (Q).

Solution:
First, we give a proof for lemma (b): we define $S_{k}=A_{k}+B_{k}$ for each integer $k$.Prove that $S_{k+1}=2 S_{k}-S_{k-1}$ for each integer $k$.

Proof for lemma (b):
We draw part of the array:

| $\cdots$ | $A_{k-1}$ | $A_{k}$ | $A_{k+1}$ | $\cdots$ |
| :--- | :--- | :--- | :--- | :--- |
| $\cdots$ | $B_{k-1}$ | $B_{k}$ | $B_{k+1}$ | $\cdots$ |

Then

$$
\begin{aligned}
3 S_{k} & =3 A_{k}+3 B_{k} \\
& =3\left(\frac{A_{k-1}+B_{k}+A_{k+1}}{3}\right)+3\left(\frac{B_{k-1}+A_{k}+B_{k+1}}{3}\right) \\
& =A_{k-1}+B_{k}+A_{k+1}+B_{k-1}+A_{k}+B_{k+1} \\
& =\left(A_{k-1}+B_{k-1}\right)+\left(A_{k}+B_{k}\right)+\left(A_{k+1}+B_{k+1}\right) \\
& =S_{k-1}+S_{k}+S_{k+1}
\end{aligned}
$$

Since $3 S_{k}=S_{k-1}+S_{k}+S_{k+1}$, then $S_{k+1}=2 S_{k}-S_{k-1}$, which finish the proof of lemma (b).


Proof of statement (P)

Suppose that all of the entries in the array are positive integers.

Assume that not all of the entries in the array are equal.

Since all of the entries are positive integers, there must be a minimum entry. Let $m$ be the minimum of all of the entries in the array.

Choose an entry in the array equal to $m$, say $A_{r}=m$ for some integer $r$. The same argument can be applied with $B_{r}=m$ if there are no entries equal to $m$ in the top row.

If not all of the entries $A_{j}$ are equal to $m$, then by moving one direction or the other along the row we will get to some point where $A_{t}=m$ for some integer $t$ but one of its neighbours is not equal to $m$. (If this were not to happen, then all of the entries in both directions would be equal to $m$.)

If all of the entries $A_{j}$ are equal to $m$, then since not all of the entries in the array are equal to $m$, then there will be an entry $B_{t}$ which is not equal to $m$.

In other words, since not all of the entries in the array are equal, then there exists an integer $t$ for which $A_{t}=m$ and not all of $A_{t-1}, A_{t+1}, B_{t}$ are equal to $m$.

But $3 m=3 A_{t}$ and $3 A_{t}=A_{t-1}+B_{t}+A_{t+1}$ so $3 m=A_{t-1}+B_{t}+A_{t+1}$.

Since not all of $A_{t-1}, B_{t}$ and $A_{t+1}$ are equal to $m$ and each is at least $m$, then one of these entries will be greater than $m$.

This means that $A_{t-1}+B_{t}+A_{t+1} \geq m+m+(m+1)=3 m+1>3 m$, which is a contradiction.

Therefore our assumption that not all of the entries are equal must be false, which means that all of the entries are equal, which proves statement $(\mathrm{P})$.

Proof of statement (Q)

Suppose that all of the entries are positive real numbers.

Assume that not all of the entries in the array are equal.

As in (b), define $S_{k}=A_{k}+B_{k}$ for each integer $k$.

Also, define $D_{k}=A_{k}-B_{k}$ for each integer $k$.

Step 1: Prove that the numbers $S_{k}$ form an arithmetic sequence

From (b), $S_{k+1}=2 S_{k}-S_{k-1}$.

Re-arranging, we see $S_{k+1}-S_{k}=S_{k}-S_{k-1}$ for each integer $k$, which means that the differences between consecutive pairs of terms are equal.

Since this is true for all integers $k$, then the difference between each pair of consecutive



terms through the whole sequence is constant, which means that the sequence is an arithmetic sequence.

Step 2: Prove that $S_{k}$ is constant

Suppose that $S_{0}=c$. Since $A_{0}>0$ and $B_{0}>0$, then $S_{0}=c>0$.

Since the terms $S_{k}$ form an arithmetic sequence, then the sequence is either constant, increasing or decreasing.

If the sequence of terms $S_{k}$ is increasing, then the common difference $d=S_{1}-S_{0}$ is positive.

Note that $S_{-1}=c-d, S_{-2}=c-2 d$, and so on.

Since $c$ and $d$ are constant, then if we move far enough back along the sequence, eventually $S_{t}$ will be negative for some integer $t$. This is a contradiction since $A_{t}>0$ and $B_{t}>0$ and $S_{t}=A_{t}+B_{t}$.

Thus, the sequence cannot be increasing.

If the sequence of terms $S_{k}$ is decreasing, then the common difference $d=S_{1}-S_{0}$ is negative.

Note that $S_{1}=c+d, S_{2}=c+2 d$, and so on.

Since $c$ and $d$ are constant, then if we move far along the sequence, eventually $S_{t}$ will be negative for some integer $t$. This is also a contradiction since $A_{t}>0$ and $B_{t}>0$ and $S_{t}=A_{t}+B_{t}$.

Thus, the sequence cannot be decreasing.

Therefore, since all of the entries are positive and the sequence $S_{k}$ is arithmetic, then $S_{k}$ is constant, say $S_{k}=c>0$ for all integers $k$.

Step 3: Determine range of possible values for $D_{k}$

We note that $S_{k}=A_{k}+B_{k}=c$ for all integers $k$ and $A_{k}>0$ and $B_{k}>0$.

Since $A_{k}>0$, then $B_{k}=S_{k}-A_{k}=c-A_{k}<c$.

Similarly, $A_{k}<c$.

Therefore, $0<A_{k}<c$ and $0<B_{k}<c$.

Since $D_{k}=A_{k}-B_{k}$, then $D_{k}<c-0=c$ and $D_{k}>0-c=-c$.

In other words, $-c<D_{k}<c$.

Step 4: $D_{k+1}=4 D_{k}-D_{k-1}$

Using a similar approach in our proof of lemma (b)

$$
\begin{aligned}
3 D_{k} & =3 A_{k}-3 B_{k} \\
3 D_{k} & =\left(A_{k-1}+B_{k}+A_{k+1}\right)-\left(B_{k-1}+A_{k}+B_{k+1}\right) \\
3 D_{k} & =\left(A_{k+1}-B_{k+1}\right)+\left(A_{k-1}-B_{k-1}\right)-\left(A_{k}-B_{k}\right) \\
3 D_{k} & =D_{k+1}+D_{k-1}-D_{k} \\
4 D_{k}-D_{k-1} & =D_{k+1}
\end{aligned}
$$

as required.

Step 5: Final contradiction 

$\text { We want to show that } D_{k}=0$ for all integers $k$.

This will show that $A_{k}=B_{k}$ for all integers $k$.

Since $S_{k}=A_{k}+B_{k}=c$ for all integers $k$, then this would show that $A_{k}=B_{k}=\frac{1}{2} c$ for all integers $k$, meaning that all entries in the array are equal.

Suppose that $D_{k} \neq 0$ for some integer $k$.

We may assume that $D_{0} \neq 0$. (If $D_{0}=0$, then because the array is infinite in both directions, we can shift the numbering of the array so that a column where $D_{k} \neq 0$ is



labelled column 0.)

Thus, $D_{0}>0$ or $D_{0}<0$.

We may assume that $D_{0}>0$. (If $D_{0}<0$, then we can switch the bottom and top rows of the array so that $D_{0}$ becomes positive.)

Suppose that $D_{1} \geq D_{0}>0$.

Then $D_{2}=4 D_{1}-D_{0} \geq 4 D_{1}-D_{1}=3 D_{1}$. Since $D_{1}>0$, this also means that $D_{2}>D_{1}>0$.

Similarly, $D_{3}=4 D_{2}-D_{1} \geq 4 D_{2}-D_{2}=3 D_{2}>D_{2}>0$. Since $D_{2} \geq 3 D_{1}$, then $D_{3} \geq 9 D_{1}$. Continuing in this way, we see that $D_{4} \geq 27 D_{1}$ and $D_{5} \geq 81 D_{1}$ and so on, with $D_{k} \geq 3^{k-1} D_{1}$ for each positive integer $k \geq 2$. Since the value of $D_{1}$ is a fixed positive real number and $D_{k}<c$ for all integers $k$, this is a contradiction, because the sequence of values $3^{k-1}$ grows without bound.

The other possibility is that $D_{1}<D_{0}$.

Here, we re-arrange $D_{k+1}=4 D_{k}-D_{k-1}$ to obtain $D_{k-1}=4 D_{k}-D_{k+1}$.

Thus, $D_{-1}=4 D_{0}-D_{1}>4 D_{0}-D_{0}=3 D_{0}>D_{0}>0$.

Extending this using a similar method, we see that $D_{-j}>3^{j} D_{0}$ for all positive integers $j$ which will lead to the same contradiction as above.

Therefore, a contradiction is obtained in all cases and so it cannot be the case that $D_{k} \neq 0$ for some integer $k$.

Since $D_{k}=0$ and $S_{k}=c$ for all integers $k$, then $A_{k}=B_{k}=\frac{1}{2} c$ for all integers $k$, which means that all entries in the array are equal.