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Take any two numbers between $0$ and $1$. Prove that the sum of the numbers is always less than one plus their product.
That is, if $0< x< 1$ and $0< y< 1$ then prove
$$x+y< 1+xy$$
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?
Two cubes, each with integral side lengths, have a combined volume equal to the total of the lengths of their edges. How big are the cubes? [If you find a result by 'trial and error' you'll need to prove you have found all possible solutions.]