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
re: when x is odd - when x is even by Brian Smith)
"Then each of p-q and p+q are powers of 2."
p+q=2^m
p-q=2^n
p=(2^n-1) * (2^m-n + 1)
(2^n-1) = 1 so n-1=0 and n=1.
That makes p=2^m-1 + 1 and q=2^m-1 - 1.
Say (m-1) is odd. Then p is divisible by 3 and thus p=3. But p>q=odd prime so this case is impossible.
Then m-1 is even and it's q that's divisible by 3 and = 3.
The rest of your solution follows.
Nice collaborating with you.
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Posted by xdog
on 2022-06-06 14:14:44 |
re(2): when x is odd - when x is even
