Each of p and q is a
prime number and each of x and y is a
positive integer, with x greater than 1.
Find quadruplet(s) (p, q, x, y) that satisfy this equation:
px - qx = 2y
providing adequate reasons as to why there are no further solutions.
*** Adapted from a problem which appeared at the Brazilian Mathematical Olympiad in 1997.
(In reply to
when x is odd by xdog)
So x must be even.
Let x=2z. Then (p^2)^z - (q^2)^z = 2^y.
But this factors into (p^2 - q^2)*(p^x-2 + q^2p^x-4 + . . . + q^x-2) = 2^y.
The important part is to realize p^2-q^2 must be a power of 2. Then each of p-q and p+q are powers of 2.
Assume that p and q are both larger than 3. Then each one must be of the form 6k+1 or 6k-1.
If both p and q are 6k+1 or both of the form 6k-1 then p-q will be a multiple of 6. This contradicts the need for p-q to be a power of 2.
If exactly one of p and q is of the form 6k+1 and the other is of the form 6k-1, then p+q will be a multiple of 6. This contradicts the need for p+q to be a power of 2.
Then one of p and q must be 3, but since q is smaller than p we must have q=3.
So now p+3 and p-3 are powers of 2. Which means we need a pair of powers of 2 that differ by 6. 8,2 is the only such pair. That makes p=5.
So we are now down to 5^x - 3^x = 25^z - 9^z = 2^y
z=1 implies p=5, q=3, x=2, y=4 is a solution.
Now we just need to polish off the last bit to show z>1 has no solutions.
re: when x is odd - when x is even
