### Upsetting Pitagoras

Find the smallest integer solution to the equation 1/x^2 + 1/y^2 = 1/z^2

### Euclid's Algorithm I

How can we solve equations like 13x + 29y = 42 or 2x +4y = 13 with the solutions x and y being integers? Read this article to find out.

### What's a Group?

Explore the properties of some groups such as: The set of all real numbers excluding -1 together with the operation x*y = xy + x + y. Find the identity and the inverse of the element x.

# Eyes Down

##### Age 16 to 18 Challenge Level:

Thank you to Alan of Madras College for this solution.

If $x$ is a real number then $x = a + b$ where $a$ is an integer and $b$ is a real number such that $0 \leq b < 1$. Here $a$ is the integer part of $x$ and we write $a = [x]$. We have to consider whether $[2x]$; $2[x]$ and $[x + 1/2 ] + [x - 1/2 ]$ can ever be equal and whether they can take three different values.

If $1/2 \leq b < 1$ then $[2x]= 2a + 1$.

If $0 \leq b < 1/2$ then $[2x]= 2a$.

For any $b$, $2[x] = 2a$.

If $1/2 \leq b < 1$ then $[x+ 1/2 ] = a + 1$ and $[x - 1/2 ] = a$ and so $[x + 1/2 ] + [x - 1/2 ] = 2a + 1$.
If $0 \leq b < 1/2$ then $[x+ 1/2 ] = a$ and $[x - 1/2 ] = a - 1$ and so $[x + 1/2 ] + [x - 1/2 ] = 2a - 1$.

$\bullet$ Case 1: $\; 0 \leq b < 1/2$

$[2x]= 2a = 2[x]$

but $[2x] \neq [x + 1/2 ] + [x - 1/2 ]$.

$\bullet$ Case 2: $\; 1/2 \leq b < 1$

$[2x]= 2a + 1 = [x + 1/2 ] + [x - 1/2]$

but $[2x] \neq 2[x]$.

Hence it is impossible for all of $[2x]$; $2[x]$ and $[x + 1/2 ] + [x - 1/2 ]$ to be equal but they can never take three different values.