X is a positive integer > 1 and, P is a prime number.
Determine all possible pairs (X, P) such that
P^{X} + 144 is a perfect square.
P^X + 144 = k^2
P^X = (k+12)(k12)
Let
k+12 = P^A  (1)
k12 = P^B  (2)
Subtract
P^A  P^B = 24
P^B(P^(AB)1) = 24
P^B(P^(AB)1) = 2^3 * 3  (3)
This implies
1) P is a divisor of 24 => P = 2 or 3 (B is not 0)
2) B = 0
Case 1)
i) P = 2
From (3) B = 1, 2 or 3
From (2) k = P^B + 12
=> k = 14, 16 or 20
=> k+12 = 26, 28 or 32
Only 32 is a power of 2 => X = A+B = 5+3= 8 (8,2)
ii) P = 3
From (3) B = 1
From (2) k = P^B + 12 = 3+12 = 15
=> k+12 = 27 = 3^3
=> X = A+B = 3+1 = 4 (4,3)
Case 2)
B = 0
From (2) k = 1+12 = 13
k+12 = 25 = 5^2
X = A+B = 2+0 = 2 (2,5)
Only Solutions are (8,2), (4,3) and (2,5)

Posted by Praneeth
on 20100113 03:28:30 