### Shades of Fermat's Last Theorem

The familiar Pythagorean 3-4-5 triple gives one solution to (x-1)^n + x^n = (x+1)^n so what about other solutions for x an integer and n= 2, 3, 4 or 5?

### Exhaustion

Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2

### Code to Zero

Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135.

# Inner Equality

##### Age 16 to 18 ShortChallenge Level
Well done to Amrit, Adithya, Daven and Sergio who all sent in solutions to this problem. Here are the first four inequalities:

$$10 < a+ b- c - d < 20$$
$$0 < a- c < 10$$
$$-10 < a - c + d - b < 10$$
$$0 < abcd < 625$$
Aditha's solution explains how to get each of them, you can read the pdf

Amrit explained how to work out the fifth inequality:

For the last inequality, we need to prove the AM-GM inequality
$\frac{a+c}{2}>\sqrt{ac}$
$a+c>2\sqrt{ac}$
$a^2+2ac+c^2>4ac$
$a^2-2ac+c^2>0$
$a-c)^2>0$
A number squared is always greater than 0 unless the number is 0 or in this case if a=c
Plugging in a=c into the last inequality, we have
$\frac{|a|+|c|}{2}-\sqrt|ac|>0$
Looking at the AM-GM inequality, we want a and c to be as far apart as
possible. So |a| has to be 5 and |c| has to be 0 or vice versa.
Applying this, we have $\frac{|a|+|c|}{2}-\sqrt|ac|<\frac{5}{2}$