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I am studying the properties of the following equation and am trying to show that there is a contradiction. I have that $$x=\Big[\frac{2q+1}{2m}\Big]\sum_{i=1}^kb_i$$ Here all $b_i$ are positive, nonzero and odd and $k$ is odd (so summing an odd number of odd terms, implying an odd result). Also, $q$ and $m$ are nonzero, positive integers. It is known that $b_k=x$. It is also known that $x$ is a positive integer and is odd. I am wondering if I can prove that the above equality cannot hold because it would imply a fractional result for $x$, which would contradict it being an integer. However, this approach only works if I don't solve for $x$ entirely, that is, it only works by not acknowledging the $b_k$ term and solving for $x$ from there. Is such an approach (not solving for $x$ in the final term of the sum) logically sound? Or is it necessary for me to solve for $x$ in its entirety and find a different route to contradiction? If so, could anyone guide me in the correct direction for such a different approach? Thanks!

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I thought I should make this more clear. I have gotten to this point in the problem: in the sum above I have that $$x=\Big[\frac{2q+1}{2m}\Big](b_1+b_2+\cdots+b_{k-1}+x) $$ I am given that $x$ is an odd, positive integer. However, the fraction preceding the sum implies $x$ is fractional (I also know the sum, including $x$ as a term, is odd). However, if I solve for $x$ completely, thereby removing $x$ from the sum, I find that my method of contraction fails, as it implies that $x$ could be odd.

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    $\begingroup$ $b_k=x$ for all $k$ or just for some unique $k$? $\endgroup$ – Masacroso Sep 15 '16 at 1:30
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    $\begingroup$ @Masacroso By $b_k=x$ I mean the last term in $\sum_{i=1}^kb_i$ is equal to $x$. All other $b_i$ are not equal to $x$. $\endgroup$ – Thomas Myer Sep 15 '16 at 2:00
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    $\begingroup$ @ Masacroso I've edited the post, hopefully its a little clearer. $\endgroup$ – Thomas Myer Sep 15 '16 at 2:22
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    $\begingroup$ @Masacroso $x$ is a fixed value---one that we know to be an odd integer and certainly greater than zero. $x$ cannot be changed. $\endgroup$ – Thomas Myer Sep 15 '16 at 2:39
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    $\begingroup$ I dont know exactly what you want here... but it seems easy to show that the RHS is not an integer. By contradiction you can assume that the RHS is an integer, and then $2m$ must divide $2q+1$ and $\sum b_j$ due to the fundamental theorem of arithmetic cause both are integer, what cannot happen. You dont need to solve for $x$, this is not necessary. $\endgroup$ – Masacroso Sep 15 '16 at 2:48
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Expanding a little bit my comment, my idea to solve this by contradiction will be something like:

1) Assume that $x\in\Bbb Z$

2) Show that $2q+1\in\Bbb Z$ and $\sum b_j\in \Bbb Z$

3) Then show that by the fundamental theorem of arithmetic $2m$ must divide $2q+1$ or $\sum b_j$.

4) Show that the OR statement cannot be possible, i.e. $2m$ cannot divide $2q+1$ and neither $\sum b_j$.

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