by quintic » Sun Apr 16, 2006 3:21 am
I may be missing a simpler approach, but here is an outline of a proof...
Writing p = 2[sup]r[/sup]a and q = 2[sup]s[/sup]b for some non-negative integers r and s, and odd integers a and b, we must have r = s, for otherwise, if (wlog) r < s, a[sup]23[/sup] + (2[sup]s-r[/sup]b)[sup]23[/sup] would be odd and greater than 1, a contradiction.
Hence we have a[sup]23[/sup] + b[sup]23[/sup] = 2[sup]n[/sup], where n is a positive integer.
Assume not both a and b are equal to 1.
If a[sup]23[/sup] + b[sup]23[/sup] = 2[sup]n[/sup], then n > 23, and
A = a + b = 2[sup]m[/sup], for some positive integer m > 1, and
B = a[sup]22[/sup] - a[sup]21[/sup]b + a[sup]20[/sup]b[sup]2[/sup] - a[sup]19[/sup]b[sup]3[/sup] + ... + b[sup]22[/sup] = 2[sup]n-m[/sup], and clearly n - m > 1.
As a + b is divisible by 4, a = +/- 1, b = -/+ 1 (mod 4). But then B = -1 (mod 4), a contradiction.
Hence a = b = 1, implying p = q = 2[sup]r[/sup].