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?
This is the solution sent in by Yatir Halevi. Thanks Yatir. A
correct solution was also received from Andrei Lazanu.
Let's say we want to find the square of $a$
We know that $a^2 = a^2-b^2+b^2 = (a+b)\times(a-b)+b^2$and for
every a, we can pick a certain b that will make the
calculation$a^2$ as easy as possible.
So, if$a$ is a number that ends with a 5: it can be written as
$$a=10\times q +
So $a^2$is equal to $q(q+1)$ plus two zeros after it $(10^2)$ that
are "stolen" by the 25 that is added on.