It can be shown easily that (x; y; z) = (2(n ¡ 1); 0; n ¡ 1) is a solution for all natural number n: Now, if
x; y; z > 0; we have 22z ¡ 2x = 7y2: Then, 2x(22z¡x ¡ 1) = 7y2: Hence, x is even so 7y2 ´ 4z ¡ 2x ´ 0 (mod 3). It
follows that y is divisible by three, i.e. y = 3k for some k 2 N. Letting y = 3k; we obtain 2x(22z¡x ¡ 1) = 63k2;
implying 2x = k2 and 22z¡x ¡ 1 = 63: The solution to 2x = k2 is then given by x = 2(n ¡ 1) and k = 2n¡1: For
22z¡x ¡1 = 63 we have the solution 2z ¡x = 6 or x = 2(z ¡3): Furthermore, we see that 2(n¡1) = 2(z ¡3); that
is z = n + 2: This completes the proof of the theorem.