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                                                        (1)

From (1):                      2^x = y²              (mod 3)
which gives                    2^x = 0 or 1         (mod 3)
Consequently,                x is even           since 2^x = 2 (mod 3) if x is odd
so let                            x = 2n   (n >= 1)
then (1) becomes           2²ⁿ - y² = 3p
                                    (2ⁿ - y)(2ⁿ + y) = 3p                                           (2)

Now, noting that the RHS has two prime factors, 3 and p, and that  the brackets are in order of magnitude:

If p = 2, then 2ⁿ - y = 2  and    2ⁿ + y = 3,  giving 2ⁿ = 5/2, which is not possible.

So, p >= 3, and (2) then has the following two possible outcomes:

2ⁿ - y = 1   and   2ⁿ + y = 3p,   giving   2ⁿ = (3p + 1)/2,  p = (2^{n + 1} - 1)/3   (A)  
or
2ⁿ - y = 3   and   2ⁿ + y = p,     giving   2ⁿ = (p + 3)/2,    p = 2^{n + 1} - 3        (B)

We need A and B to have distinct solutions (n1, y1) and (n2, y2) respectively, for a common value of p. Thus

from (A) and (B)                                    (2^{n1 + 1} - 1)/3 = 2^{n2 + 1} - 3
which gives                                2^{n1 + 1} = 3(2^{n2 + 1}) - 8
Dividing by 8                             3(2^{n2 - 2}) = 2^{n1 - 2} + 1                                (3)

Substituting p >= 3 in (A), gives 2ⁿ¹ >= 5; therefore n1 > 2 and RHS of (3) is odd.
Substituting p >= 3 in (B), gives 2ⁿ² >= 3; therefore n2 > 1.
Since the LHS of (3) must be odd, n2 - 2 < 1, giving n2 < 3; therefore n2 = 2 and n1 = 3, which, when used in (A) and (B) respectively, both give p = 5.

So p = 5 is the only solution, with (x, y) being (6, 7) and (4, 1).

Edited on October 24, 2009, 9:30 pm

Posted by Harry on 2009-10-24 21:27:59