<|im_start|>system
You are a helpful AI assistant.<|im_end|>
<|im_start|>user
How do I make a table in MathJax?<|im_end|>
<|im_start|>assistant
You can make a table in MathJax using the following syntax:
``` 
\begin{array}{column specification}
cell & cell & cell \\
cell & cell & cell \\
cell & cell & cell
\end{array}
``` 
The column specification defines the layout of the table, where each column is separated by & and each row is separated by \\. For example, the following code creates a simple 2x2 table with centered columns:
``` 
\begin{array}{c|c}
a & b \\
\hline
c & d
\end{array}
``` 
This produces the following table:
```
a|b
---
c|d 
```
You can adjust the column specification to create tables with different layouts and alignment options. For more information and examples, you can refer to the MathJax documentation on tables: https://docs.mathjax.org/en/latest/input/tex/arrays.html.<|im_end|>
<|im_start|>user
That link gives me a 404. Do you have another link that might work?<|im_end|>
<|im_start|>assistant
I apologize for the confusion. The MathJax documentation on tables seems to have been moved or removed. Here is an alternative resource that provides information on how to create tables in MathJax:

https://en.wikibooks.org/wiki/LaTeX/Tables#Using_the_array_environment

This resource provides an overview of the array environment in LaTeX, which is used to create tables in MathJax. The syntax is similar to what I provided earlier, but this resource includes additional information on how to specify column alignments and formatting options.

I hope this helps!<|im_end|>
<|im_start|>user
Thanks for your answer, now I want that you solve the following problem: 

Let a and b be positive integers such that ab + 1 divides a² + b². Show that

a² + b² / ab + 1

is the square of an integer.<|im_end|>
<|im_start|>assistant
To show that a² + b² / ab + 1 is the square of an integer, we need to find an integer k such that:

a² + b² / ab + 1 = k²

To start, we can simplify the expression on the left-hand side by multiplying both the numerator and denominator by a - b:

(a² + b²)(a - b) / (ab + 1)(a - b) = k²

Expanding the numerator on the left-hand side and canceling out the common factor of a - b in both the numerator and denominator, we get:

(a³ - b³) + ab(a - b) / (a²b - ab²) + 1 = k²

Since ab + 1 divides a² + b², we can write:

a² + b² = n(ab + 1)

where n is some positive integer. Substituting this expression into the numerator of the left-hand side, we get:

(a³ - b³) + ab(a - b) = a³ - b³ + ab(a - b) = (a - b)(a² - ab + b²) = (a - b)n

Substituting this expression back into our original equation and simplifying, we get:

(a - b)n / ab - 1 = k²

Multiplying both sides by ab - 1, we get:

(a - b)n = k²(ab - 1)

Since a, b, and n are positive integers, we have:

a - b > 0 and ab - 1 > 0

Therefore, k² must also be a positive integer, and we can write:

k² = (a - b)n / (ab - 1)

Now we need to show that k is an integer. To do this, we will use the fact that ab + 1 divides a² + b². This can be written as:

a² + b² = m(ab + 1)

where m is some positive integer. Substituting this expression into the original equation, we get:

m(ab + 1) / ab + 1 = k²

Multiplying both sides by ab + 1, we get:

m(ab + 1) = k²(ab + 1)

Expanding the left-hand side and simplifying, we get:

mab + m = k²ab + k²

Dividing both sides by ab, we get:

m/a + m/b = k² + m/(ab)

Since m/(ab) is a positive integer and m/a and m/b are both positive integers, k² must also be a positive integer, and k is an integer.

Therefore, we have shown that a² + b² / ab + 1 is the square of an integer.<|im_end|>