Determine all possible values of a positive integer N ≥ 3, such that ^NC₂ – 1 is a prime power.

Note: ^NC₂ represents N choose 2.

N choose 2 = n(n-1)/2

subtract 1 and factor to get
.5n - .5n - 1 = (n-2)(n+1)/2

(n-2) and (n+1) differ by 3.  The 1/2 will divide into whichever factor is even giving either [1] (.5n-1)(n+1) or [2] (n-2)(.5n+.5)

This leaves two possibilities:
(A) One of the factors is 1 and the other is prime. 
(B) Both factors are powers of 3.

Checking possibilities for (A1)
If n=4 then (.5n-1)=1 and (n+1)=5  [4 is a solution]
Only n=0 makes (n+1)=1

Checking possibilities for (A2)
If n=3 then (n-2)=1 and (.5n+.5)=2 [3 is a solution]
Only n=1 makes (.5n+.5)=1

Possibilities for (B1)
.5n-1 = 3^x and n+1=3^y
Which means
2*3^x+3 = 3^y
The only possibility is 2*3^1+3 = 3^2 = 9
.5n-1 = 3 and n+1 = 9
n = 8 [8 is a solution]

Possibilities for (B2)
n-2=3^x and .5n+.5=3^y
Which means
3^x = 2*3^y - 3 
The only possibility is 3^1 = 2*3^1 - 3 = 3
n-2=3
n = 5 [5 is a solution]

So the only four solutions are n={3,4,5,8}

Posted by Jer on 2010-04-23 16:28:43