Case 1: If y is odd, say y = 2k-1 then 2^(2k-1) = 3w + 2 where
w is an integer. But 3^x -17 = 3(3^(x-1)-6) + 1 = 3v + 1 where v
is an integer. So this case has no solutions.
Case 2: If y=4k-2 and x is odd, say x=2j-1, then 17 + 2^(4k-2)=
10w + 1. But 3^(2j-1) = 10v + 3 or 10v + 7 so this case has no solution.
Case 3: If y=4k and x=4j-1, then 17+2^(4k) = 10w + 3. But
3^(4j-1) = 10v + 7 so this case also yields no solutions.
Case 4: If y=4k and x = 4j-3 then 17+2^(4k)=17w + 16 or
17w + 1. But 3^(4j-3)=17v + 3 or 17v + 5 or 17v + 14 or
17v + 12 so again, this case yields no solutions.
Case 5: Finally, if x and y are both even, say y=2k and x=2j,
then 3^x - 2^y = (3^j)^2 - (2^k)^2 = (3^j-2^k)(3^j+2^k) =
17. Since 17 is prime, this forces 3^j + 2^k = 17 so only j=1 and
j=2 need be considered. Only j=2 yields a viable solution forcing
k=3 and in turn the unique solution x=4 and y=6.
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Posted by Dennis
on 2006-09-14 10:44:09 |
a solution
