edit

11^2 + 12^1 = 133

11n+3 + 12^2n+3

But then I get stuck!? Any ideas?

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- Nov 20th 2007, 02:13 PMJimboboboProof by induction! Urgh! :(
edit

11^2 + 12^1 = 133

11n+3 + 12^2n+3

But then I get stuck!? Any ideas? - Nov 20th 2007, 02:42 PMPlato
Say it is true for K>1, $\displaystyle \left( {11} \right)^{K + 2} + \left( {12} \right)^{2K + 1} = 133j$.

Look at the case of K+1.

$\displaystyle \left( {11} \right)^{\left( {K + 1} \right) + 2} + \left( {12} \right)^{2\left( {K + 1} \right) + 1}$

$\displaystyle = \left( {11} \right)^{K + 3} + 11\left( {12} \right)^{2K + 1} - 11\left( {12} \right)^{2K + 1} + \left( {12} \right)^{2K + 3} $

$\displaystyle \begin{array}{l}

= 11\left[ {\left( {11} \right)^{K + 2} + \left( {12} \right)^{2K + 1} } \right] + \left( {12} \right)^{2K + 1} \left[ { - 11 + \left( {12} \right)^2 } \right] \\

= 11\left[ {133j} \right] + \left( {12} \right)^{2K + 1} \left[ {133} \right] \\

\end{array}

$ - Nov 21st 2007, 08:30 AMtopsquark
Plato either wins the award for most psychic Math Helper, or Jimbobobo edited his original question to the point where it is meaningless. I'm curious as to which?

-Dan - Nov 21st 2007, 08:43 AMPlato
The original question was edited.