Suppose there exists a solution. Then by the Well Ordering principle, there exists a solution for which A+B is the minimum, as A+B is bounded below by 0.
In this "minimal" solution, (36A+B) and (36B+A) must both be greater than 2 and thus, also powers of 2 if their product is a power of 2. This implies C is greater than 2 but also that A and B must be even, as if A (or B) were odd, then 36B+A (or 36A+B) would be odd and not a power of 2.
However, since both A and B are even and C is greater than 2, then A'=A/2, B'=B/2, and C'=C-2 are all positive integers. So by (A,B,C) being a solution, then (36A+B)(36B+A) = 2^C.
Dividing by 2 twice on both sides gives (36(A/2)+(B/2))(36(B/2)+(A/2))=2^(C-2) or (36A'+B')(36B'+A')=2^C'.
Then A'+B'=A/2+B/2 is less than A+B, which contradicts it being the minimal solution. Thus, the assumption that there is a solution must be false.
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Posted by Gamer
on 2008-07-12 17:33:39 |
Contradiction solution (quicker but trickier)
