### 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 18 Challenge Level:

Try a proof by contradiction and use the Triangle Inequality which says that a triangle can be constructed with three given segments for sides if and only if the sum of the lengths of any two exceeds the length of the third. (For example the lengths $2$, $3$ and $7$ cannot make the sides of a triangle because $2+3 < 7$.)

One more hint, one of the edges of the tetrahedron must be the longest and, without loss of generality, you can label this edge $AB$. Now, if you are using a proof by contradiction, what can you say about the 3 edges meeting at $A$ and similarly about the three edges meeting at $B$?