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)(k-12)

Let 

k+12 = P^A -- (1)

k-12 = P^B -- (2)


Subtract

P^A - P^B = 24

P^B(P^(A-B)-1) = 24

P^B(P^(A-B)-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 2010-01-13 03:28:30