1. Working in base ab throughout and ignoring n=0;
2. Let b be the larger of a,b with a,b, relatively prime.
3. First multiply b^n by a^n, less the first digit of a^n, x.
4. Next multiply every digit of b^n except the first, again x, by x.
5. The sum of 3 and 4 is C followed by a lot of zeros:
C(ab)^(n1) =(ab)^nx^2(ab)^(n1)
6. Now we have x^2+C=10=ab
7. The first digit of a^n is some power of (ab), say x(ab)^k; the first digit of b is similarly x(ab)^l
8. Since a^n and b^n start with the same digit, we can also say that
b^n/a^n is (roughly) some power of (ab), to be precise
(ab)^l/(ab)^k or (ab)^(lk); so x approximates to (ab)^.5)
and C is less than x^2 since (x+1)^2>2x^2 only for x = {0,1,2}
10. Let a = 2, b=7; we seek some x such that x^2+C = 10, giving {(1,D)(4,A)(9,5)}
11. C is less than x^2 iff x= 3.
12. The above doesn't guarantee solutions, it just states what form they will take if they exist.
13. However, 2^17= 3c8,3a8, 7^17 = 3,839,d52,0c8,c06,1a7 is a possible solution.
A truly remarkable result  wholly counterintuitive.
Edited on November 20, 2010, 2:34 pm

Posted by broll
on 20101120 11:21:37 