[Guest]

Determine all possible values of a positive integer X >= 3, such that ^XC₂ – 1 is a prime power.

Note: ^XC₂ represents X choose 2.

[Guest]

^xC₂ - 1 = (x(x-1)/2) - 1 = (x+1)(x-2)/2

Now this must be a power of a prime number

Since, the difference in terms x+1 and x-2 is only 3 and since one of them must be a multiple of 2,

Case 1)

lets say x+1 = 2*aⁿ where a is a prime number

Then x-2 = a^m

Also,

aⁿ = a^p * a^m because the product must be a power of a prime

Further,

2*aⁿ = 3 + a^m

2*a^p * a^m = 3 + a^m

Then,

a^m (2*a^p - 1) = 3

This is possible when

case a)

a = 3 and m = 1 and p = 0

This means x-2 = 1 meaning x = 3

case b)

a = 2 and m = 0 and p = 1

This means x - 2 = 3 meaning x = 5

Case 2)

x-2 = 2*aⁿ

x+1 = a^m

Then

a^m = aⁿ * a^p

Also, a^m - 2*aⁿ = 3

Then, 3 + 2*aⁿ = aⁿ * a^p

3 = aⁿ * (a^p - 2aⁿ)

This is possible when

Case a)

aⁿ = 1 a^p = 5

This means; a = 5 and n = 0 and p = 1

This gives x-2 = 2*1 ;; x = 4

Case b)

aⁿ = 3 and a^p = 7

This gives x-2 = 2*3;; x = 8

So, the only solutions for x are 3,4,5,and 8

Edited January 19, 2010 by DeeGee