Heiro: Good morning, Archimedes! I bet you 1000 drachma that you can’t answer a simple maths problem!

Archimedes: You’re on.

Heiro: Did you know that every sum of consecutive cubes is also a difference of squares?

Archimedes: Indeed, as everyone knows, each and every sum of consecutive cubes is also a difference of consecutive squares. Was that the question?

Heiro: Not so fast, Archimedes! What if the sum and difference are a prime number?

Archimedes (yawning): Unlikely, since any sum of cubes has at least two factors…

Heiro: Let’s make it a difference of consecutive cubes, then?

Archimedes: There is no reason why such a difference should not be prime, as often as occasion demands. For example, 2^3-1^3 = 7. Have I won yet?

Heiro: I'm just working up to it. What about a difference of consecutive cubes, which is not only prime, but is also a sum of consecutive squares?

Archimedes: Too easy! 61 = 5^3-4^3=5^2+6^2. Can I have my money now?

Heiro: Very well, then, here's my question – just to make it interesting, let’s also stipulate that a side of the larger cube must have at least 4 distinct prime factors – so it must be at least 2*3*5*7 = 210, say? Just how big would that cube have to be?

Archimedes: Easy again! – or is it? Wait a minute…

What is the answer to Heiro’s simple problem?

(In reply to first four by Daniel)

Well done Daniel, although to be precise Heiro was asking for the size of the cube, namely: 63,030,438,032,009,860,498,224,410,471,376,639,125. The corresponding prime is 47,513,986,677,009,248,633,982,421; clearly, Archimedes would have had no means of establishing the primality of that number, hence his perplexity.

If the solutions are set out in a table, the table exhibits a number of interesting features: a, the larger of the numbers to be cubed, is A054318 in sloane; a-1 is A105038 in sloane; b-1 is A087125 in sloane; P is always in A003215 in sloane.

Posted by broll on 2010-09-09 01:51:42