11Well, (a,b,c)=(0,1,1) is the only solution it seems.
\hspace{-16}$How Many Positive Integer Solution $\mathbf{(a,b,c)}$ are\\\\ there to the equation $\mathbf{2004^a+2005^b=2006^c}$
-
UP 0 DOWN 0 0 7
7 Answers
1very basic Diophantine trick is to check that both LHS and RHS should have same parity i.e both should be odd or both should be even
2006^{c}-2004^{a} = 2005^{b}
While LHS will always be even for positive integer a,b,c RHS will always be odd.
Hence no solutions.
:)
1@Sambit
(0,1,1) implies 2005 = 2006!
Is it something like 2=1 (fallacy)
