Determine all possible value(s) of a positive prime constant P such that the equation: 2^X + 3P = Y² has precisely two distinct solutions in (X, Y), where each of X and Y is a positive integer.

2^X + 3P = Y²  gives  2^X = Y² (mod 3)

2^X can only be 1 or 2 (mod 3) while Y² can only be 0 or 1 (mod 3),
so they must each be 1 (mod 3)

However, 2^X = 1 (mod 3) means that X must be even. Now let X = 2z, so that the original equation can be written as Y² – 2^2z = 3P and factorised to give:

(Y – 2^z)(Y + 2^z) = 3P.

Since P is prime and the brackets differ by more than 1, P = 2 and P = 3 can be ruled out, leaving only two possible cases:

(A)        Y – 2^z = 3 and Y + 2^z = P,
            which gives 2^z = (P – 3)/2 and therefore  P = 2^{z + 1} + 3
or
(B)        Y – 2^z = 1 and Y + 2^z = 3P,
            which gives 2^z = (3P – 1)/2 and therefore  P = (2^{z + 1} + 1)/3

For a single value of P to correspond to two distinct values of z (say za and zb), they must come from the different cases above. Thus:

            2^{za + 1} + 3 = (2^{zb + 1} + 1)/3, which gives     3(2^{za + 1}) + 8 = 2^{zb + 1}
Dividing by 8 gives
                                    3(2^{za – 2}) + 1 = 2^{zb – 2}   

Now, za < zb so if za – 2 >0 the left and right sides would be disparate (odd and even respectively). So the only possibility is that za = 2, from which zb = 4.
These then give the only pairs for (X, Y) as (4, 7) and (8, 17), corresponding
to P = 11.

Posted by Harry on 2010-01-26 14:55:21