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Euler's Squares

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.

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Diophantine N-tuples

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.

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There's a Limit

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?

Really Mr. Bond

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

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.

For instance if we take $a=35$, we can take $b=5$, we get $35^2=(35+5)\times(35-5)+5^2 =40\times30+25 =1200+25 =1225$.

So, if$a$ is a number that ends with a 5: it can be written as $$a=10\times q + 5a^2=(10q+5)^2=(10q+5-5)\times(10q+5+5)+25=10q(10q+10)+25=10^2q(q+1)+25$$ 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.