### Ab Surd Ity

Find the value of sqrt(2+sqrt3)-sqrt(2-sqrt3)and then of cuberoot(2+sqrt5)+cuberoot(2-sqrt5).

### Absurdity Again

What is the value of the integers a and b where sqrt(8-4sqrt3) = sqrt a - sqrt b?

### Route to Root

A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence? Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give? What happens when you try this method for fourth roots or fifth roots etc.?

# Em'power'ed

##### Age 16 to 18Challenge Level

Find the smallest numbers $a, b$, and $c$ such that: $a^2 = 2 b^3 = 3 c^5.$ What can you say about other solutions to this problem? Congratulations for your good solutions to Ella and Elizabeth, S6, Madras College and Yiwan, The Chinese High Singapore. Here is Yiwan's solution: $$a^2=2b^3=3c^5.$$ As (2,3)=1, that is 2 and 3 have no common divisor other than 1, we shall write $a$, $b$, and $c$ in terms of powers of 2 and 3. Let $c=2^p3^q$ (where p, q are integer numbers above 0). Then $$3c^5=2^{5p}3^{5q+1}=2b^3=a^2.$$ Hence $$b=2^{(5p-1)/3}3^{(5q+1)/3};$$ $$a=2^{5p/2}3^{(5q+1)/2}.$$ As a, b are all integers, it follows that $3| (5p-1)$, $3| (5q+1)$, $2| (5p)$ and $2| (5q+1)$ [using the notation $3| (5p-1)$ to mean 3 divides or is a factor of $(5p-1)$]. Obviously the solution for the smallest number is when p=2 and q=1. In this case, $c=2^2 \times 3=12$ ; $b=2^3 \times 3^2=72$ ; $a=2^5 \times 3^3=864.$ The smallest solution is $(a=864, b=72, c=12).$ For other solutions take $$p=\{2, 8, 14, 20, ... 6m + 2\}$$ where m is a positive integer and $$q=\{1, 7, 13, 19, ... 6n - 5\}$$ where n is a positive integer. If we substitute any value of $p$ and $q$ from the corresponding domain, we will get the other solutions for the equation.