Find the smallest integer solution to the equation 1/x^2 + 1/y^2 = 1/z^2
A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?
Euler found four whole numbers such that the sum of any two of the
numbers is a perfect square. Three of the numbers that he found are
a = 18530, b=65570, c=45986. Find the fourth number, x. You could
do this by trial and error, and a spreadsheet would be a good tool
for such work. Write down a+x = P^2, b+x = Q^2, c+x = R^2, and then
focus on Q^2-R^2=b-c which is known. Moreover you know that Q >
sqrtb and R > sqrtc . Use this to show that Q-R is less than or
equal to 41 . Use a spreadsheet to calculate values of Q+R , Q and
x for values of Q-R from 1 to 41 , and hence to find the value of x
for which a+x is a perfect square.
Let a(n) be the number of ways of expressing the
integer n as an ordered sum of 1's and 2's. (For example
a(4) = 5 because 4 = 2+2 = 2+1+1 = 1+2+1 = 1+1+2 =
1+1+1+1). Let b(n) be the number of ways of expressing
n as an ordered sum of integers greater than 1; for
example 4 can be written as 4 or as 2+2 so b(4) = 2. As
another example, 5 can be written as 5 or as 2+3 or as 3+2 so
The values of a(n) and b(n) for n
< 9 are given in the following table. What do you notice about
These are Fibonacci sequences and a(n) = b(
n+2) for n >= 1.
The proof follows. To count the number of ways to write
n as an ordered sum of 1's and 2's notice that any ordered
sum must end in a 1 or a 2. If it ends in a 1 then there are
a( n-1) ways of writing the previous terms and if
it ends in a 2 then there are a( n-2) ways of
forming these ordered sums. Hence the total number a(n) is
given by a( n-1) + a( n-2).
This is the recurrence relation for the Fibonacci sequence so
a(n) is a `Fib'.
For b(n), that is for representations of n as
sums of integers greater than 1, we first suppose the last term is
k. Then we have:
n = * + * + ... + k (with k >= 2)
If k=2 then the terms that come before have to add up
to ( n-2) and so there are b( n-2)
possible initial decompositions into ordered sums of integers.
If k>=3 then we subtract 1 in equation (1) to
n - 1 = * + * + ... + ( k-1) (with
k-1 >=2) (2)
This has reduced the remaining representations to the conditions
of the original definition, thus there are b(
n-1) such representations. Hence b(n) =
b( n-1) + b( n-2) giving the
As b(3) = 1 = a(1) and b(4) =
a(2), where a(n) and b(n) satisfy the
same recurrence relation when n>=2, it follows that
b( n+2) = a(n) for n>=2.