Rationals Between

What fractions can you find between the square roots of 56 and 58?

Root to Poly

Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.

Consecutive Squares

The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?

Power Countdown

Stage: 4 Challenge Level:

For making 8, everyone noticed that the only ways of doing this are $2^3$, and $64^\frac{1}{2}$, which is $4^\frac{3}{2}=2^3$, and so these are the same.

Zack, Haren and Niall from JAPS made $125$ by doing $5^3$ where $3=81^\frac{1}{4}$

Charlotte and Jai from Tytherington High School sent us the following solutions:

$16^5=1048576$
$1048576^\frac{1}{2}=1024$

$16^\frac{1}{2}=4$
$243^\frac{1}{5}=3$
$4^3=64$

or alternatively:
$243^\frac{1}{5}=3$
$16^3=4096$
$4096^\frac{1}{2}=64$

Can you see why these two methods are equivalent?

Daniel, Alex, Nick and George from Tytherington gave these solutions:

$243^\frac{1}{5}=3$
$512^\frac{1}{3}=8$
$8^2=64$

$243^\frac{1}{5}=3$
$343^\frac{1}{3}=7$
$7^2=49$

Tom from Devenport Boys gave an explanation of why the last two values are unobtainable;

$89$ is a prime number, and so its only factors are $1$ and $89$, which also means that all its powers, such as $89^2$, will only have factors $1, 89, 7921$ and so on, so it is not a power of any other numbers, and so cannot be achieved.

$216=6^3$, and there is no way of achieving 6 with the values given, with a similar factor method.