You may also like

problem icon

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.

problem icon

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.

problem icon

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?

Never Prime

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

Good solutions here came from Sam, Dave, Richard and Joe, Matthew and Ross, all at Madras College, St Andrews. Other people tested special cases but did not prove the general results.


If a 2 digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger we prove that this difference can never be prime.

Let the 2 digit number be $a$ where $a> b$. Then $$ab - ba = (10a + b) - (10b + a) = 9(a - b).$$

As $9(a - b)$ is a multiple of $9$, it is not prime.

Now let the 3 digit number be $abc$ $$abc - cba = (100a + 10b + c) - (100c + 10b + a) = 99 (a - c).$$ As $99(a - c)$ is a multiple of $99$, it is not prime.

The 4 digit number can be taken as $abcd$. $$abcd - dcba = (1000a + 100b + 10c + d) - (1000d + 100c + 10b + a) = 9(111a + 10b - 10c - 111d).$$ Again, for any 4 digit number, this difference is a multiple of 9 and so it can't be a prime number.

Similarly for 5 digit numbers: $$\eqalign { abcde - edcba &= (10,000a + 1000b + 100c +10d + e) - (10,000e + 1000d + 100c + 10b + a) \cr &= 99(101a + 10b - 10d - 101e).}$$

This number is a multiple of 99 so it will never be prime.