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.
Take any whole number q. Calculate q^2 - 1. Factorize
q^2-1 to give two factors a and b (not necessarily q+1 and q-1). Put c = a + b + 2q . Then you will find that ab+1 , bc+1 and ca+1 are all perfect squares. Prove that this method always gives three perfect squares.
The numbers a1, a2, ... an are called a Diophantine n-tuple if aras + 1 is a perfect square whenever r is not equal to s . The whole subject started with Diophantus of Alexandria who found that the rational numbers
1/16, 33/16, 68/16 and 105/16 have this property. Fermat was the first person to find a Diophantine 4-tuple with whole numbers, namely 1, 3, 8 and 120. Even now no Diophantine 5-tuple with whole numbers is known.
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you
notice when successive terms are taken? What happens to the terms
if the fraction goes on indefinitely?
Conjectures are important, and should be encouraged, but along
with a challenge to really explain why any claim might be true
We hope the problem will give students a genuine pleasure in
discerning real structure, and lead their interest on into Number
The articles on the NRICH site (see link from problem page) are
an excellent follow-on.