I can't seem to figure out the logic behind this problem.

Would this problem have something to do with mode?

I know I am missing the main point of this problem.(Headbang)

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- Jan 16th 2012, 05:34 PMVeronica1999squares
I can't seem to figure out the logic behind this problem.

Would this problem have something to do with mode?

I know I am missing the main point of this problem.(Headbang) - Jan 16th 2012, 11:58 PMprincepsRe: squares
Since $\displaystyle 2^{n-12} +49 $ is an odd number it must be perfect square also . So:

$\displaystyle x^2=2^{n-12}+49 \Rightarrow 2^{n-12}=x^2-49$

Now, if we make substitution $\displaystyle y=n-12$ we can write :

$\displaystyle 2^y=x^2-49 \Rightarrow y= \log_{2} (x^2-49)$

This equality has integer solutions only if $\displaystyle x^2-49$ is power of $\displaystyle 2$ so $\displaystyle x$ must be an odd number so we shall make substitution $\displaystyle x=2m-1$ and therefore :

$\displaystyle y=\log_{2} (4m^2-4m-48)=\log_{2}(4 \cdot(m^2-m-12))$

$\displaystyle =2+\log_{2} (m^2-4m+3m-12)=$

$\displaystyle =2+\log_{2} (m+3)(m-4)=2+\log_{2} (m+3)+\log_{2}(m-4)$

It is obvious that both $\displaystyle m+3$ and $\displaystyle m-4$ must be powers of $\displaystyle 2$ and this is possible only if $\displaystyle m=5$ , therefore :

$\displaystyle y=2+3+0 \Rightarrow y=5 \Rightarrow n-12=5 \Rightarrow n=17 .$