LogFAQs > #967642862

Well, let's see.

All square numbers end in a 0, 1, 4, 5, 6, or 9.

Cubes end in all numbers other than 2 and 8.

p^2 - p - 1 will always be odd, so the numbers we are looking to hit are 1, 3, 5, 7, and 9.

All primes (5 excluded, but it doesn't work, although 2 does) end in 1, 3, 7, or 9. p^2 will therefore always end in a 1 or a 9. p^2-p-1 will therefore end in a 9 (1), 5 (3), 1 (7), or 9 (9). All cubes ending in 3 (originating from numbers ending in 7) and 7 (from numbers ending in 3), therefore, are ineligible; in other words, all applicable cubes must originate from a number ending in 1, 5, or 9.

Let's try rephrasing this.

p(p-1)-1 = c^3
p(p-1) = c^3-1 = (c-1)(c^2+c+1)

I feel like I should be able to draw something from this, but I have no idea what. I'm a physicist, not a mathematician, but I do enjoy me a good proof.