### Upsetting Pitagoras

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

### Euler's Squares

Euler found four whole numbers such that the sum of any two of the numbers is a perfect square...

### Diophantine N-tuples

Can you explain why a sequence of operations always gives you perfect squares?

# Rudolff's Problem

##### Age 14 to 16 Challenge Level:

Congratulations to Elizabeth Whitmore and Sue Liu of Madras College, St Andrew's, Scotland, to Giulio Tiozzo, age 15, Liceo Scientifico "Galileo Ferraris", Turin, Italy, and to Vassil Vassilev, age 14, Lawnswood High School, Leeds and Jonathan Kemp, age 17, Westwood High School, Leek, England. Here is Jonathan's solution.

The problem states that a group of 20 people pay a total of £20 to see an exhibition, admission £3 for men, £2 for women and 50p for children.

Denoting the number of men as $x$, the number of women as $y$ and the number of children$z$ you have to solve two simultaneous equations:$$x + y + z = 20$$ and $$3x + 2y + {z\over 2} = 20.$$ Multiplying the second equation by 2 and subtracting the first equation to eliminate $z$ gives: $$5x + 3y =20.$$ This is a Diophantine equation as featured in this months article. It is easy to spot two solutions, namely $x = 4$ , $y = 0$, and $x = 1$, $y = 5$. In this case we have 2 simultaneous equations and 3 unknowns.

Substituting these solutions into the first two equations gives: $4 + 0 + z = 20$ therefore $z = 16$ and $12 + 0 + z/2 =20$ and again $z = 16$.

Substituting $x = 1$ and $y = 5$ gives: $1 + 5 + z = 20$ therefore $z = 14$ and $3 + 10 + z/2 = 20$ and again $z = 14$.

So the group can either consist of either 4 men, no women and 16 children, or of 1 man, 5 women and 14 children.