### 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.

# Tetra Inequalities

##### Age 16 to 18Challenge Level

This solution came from Ruth from Manchester High School for Girls. Well done Ruth!

Let the tetrahedron's vertices be $A$, $B$, $C$ and $D$ and the longest side be $AB$. If you assume that there is not a vertex where the three sides meeting at it could be the sides of a triangle, we must have $AC + AD < AB$ and $BC + BD < AB$ (otherwise the sides meeting at $A$ or $B$ could be the sides of a triangle). Therefore

$$AC + AD + BC + BD< 2 AB$$.Now since $ABC$ and $ABD$ are both triangles, we must have $AC + BC > AB$ and $AD + BD > AB$. Therefore

$$AC+ AD + BC + BD > 2 AB$$ This contradicts (1), so the initial assumption must be wrong. There is at least one vertex (one of $A$ or $B$) where the three sides meeting at it could be the sides of a triangle.