Determine all possible values of a positive integer N ≥ 3, such that
^{N}C
_{2} – 1 is a
prime power.
Note:
^{N}C
_{2} represents N
choose 2.
N choose 2 = n(n1)/2
subtract 1 and factor to get
.5n  .5n  1 = (n2)(n+1)/2
(n2) and (n+1) differ by 3. The 1/2 will divide into whichever factor is even giving either [1] (.5n1)(n+1) or [2] (n2)(.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 (.5n1)=1 and (n+1)=5 [4 is a solution]
Only n=0 makes (n+1)=1
Checking possibilities for (A2)
If n=3 then (n2)=1 and (.5n+.5)=2 [3 is a solution]
Only n=1 makes (.5n+.5)=1
Possibilities for (B1)
.5n1 = 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
.5n1 = 3 and n+1 = 9
n = 8 [8 is a solution]
Possibilities for (B2)
n2=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
n2=3
n = 5 [5 is a solution]
So the only four solutions are n={3,4,5,8}

Posted by Jer
on 20100423 16:28:43 