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^2  12^2
P^x = (k+12)(k12)
P has two powers (a and b where a+b=x) such that p^a and p^b differ by 24.
For any P, P^1  P^0 = P1 but since 25 is not prime (1,25) is not a solution.
The next smallest difference is P^2  P^1 = P(P1)
P(P1) = 24 has positive solution 5.42 meaning no prime above this will work.
Therefore the only possibilities for P are 2, 3 and 5.
2^a  2^b = 24
2^b(2^(ab)1)=24
2^b is a factor of 24 which means b can only be 0, 1, 2 or 3
Checking these only b=3 gives a solution a=5 so x=3+5=8
One solution is (8,2)
3^b(3^(ab)1) = 24
3^b is a factor of 24 which means b can only be 0 or 1
Checking these only b=1 gives a solution a=3 so x=3+1=4
Second solution is (4,3)
5^b(5^(ab)1) = 24
5^b is a factor of 24 which means b can only be 0
Checking this it turns out b=0 gives a solution a=2 so x=2+0=2
Third and final solution is (2,5)

Posted by Jer
on 20100112 14:39:18 