Current Events > Tough math question for all the 140+ IQers here.

(+) Yeah! (+)

For what prime values of p is p^2-p-1 a perfect cube? Of course deriving the answer is what's most important.

7

2

4 - 2 - 1 = 1

1*1*1 = 1

surely Error1355 has an IQ of 140

Got a 130 IQ, but did not take college mathematics. It's all Greek to me.

69

Do your own homework

I'm afraid I've had a bit to drink, and my Mathematics lecturer always said, "If you can't remember anything else, remember this: don't drive and derive".

Well, I said 'brown'...

p = 37

37 - 37 - 1
1,369 - 37 - 1
1,331

31,331 = 11

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.

My IQ is only 139 so this is too advanced for me

My iq is 148, but I didn't take anything past introduction to calc in college dude.

I don't know if I could solve this tbh.

Either way though, asking someone a particular question in math because they have a high iq, is like saying someone is stupid because they don't know a specific subject they weren't taught.

A

A truly intelligent person understands that it's not worth their time to figure this out.