On a conjecture that two intersections are at $\big(\pm\frac{S}m, S\big)$ (about linear equations) Suppose we have two linear equations, $y=mx$ and $y=-mx$ , inserted on the same plane for $m>0$. These equations are going to have an intersection at $(0,0)$. Now, let $$\begin{align}S&=1+\sum_{i=2}^{m-1}\left(i-\frac12\right)\\ &\stackrel{\small\text{or}}{=}1+\left\{\bigg(2-\frac12\bigg)+\bigg(3-\frac12\bigg)+\cdots+\bigg(m-1-\frac12\bigg)\right\}.\end{align}$$
If we insert the equation, $y={m^2+1 \over 2}$, on the selfsame plane, then this horizontal line will pass through both the aforementioned linear equations, thus forming two more intersections.


Conjecture:

The two intersections are at $\big(\pm\frac{S}m, S \big)$, such that the positive $x$ coordinate is on the RHS of the $y$ axis, and the negative $x$ coordinate is on the LHS of the $y$ axis.

This is my own conjecture actually... and I can't seem to prove it. The following is my attempt, in which I did make some progress... but it is not a proof (...yet).

Attempt: $$\begin{align}S&=1+\left\{\bigg(2-\frac12\bigg)+\bigg(3-\frac12\bigg)+\cdots+\bigg(m-1-\frac12\bigg)\right\}\tag{given} \\ \\ 2\cdot S&=2+(4-1)+(6-1)+\cdots+(2m-2-1)\tag{$\times 2$} \\ &=\require{cancel}{\cancel2}+(4-1)+(6-1)+\cdots+(2m\cancel{-2}-1)\tag{$2-2=0$} \\ &= (4-1)+(6-1)+\cdots+(2m-1)\tag{simplifying} \\ &=\sum_{i=2}^m(2i-1).\tag{$\ast$}\end{align}$$
Here, we can use the following lemma to enable us to write Eq. $(\ast)$ in closed form: $$\sum_{j=1}^n(2j-1)=n^2.$$ The proof of this lemma can be found in Intermediate Mathematics: Proof by Induction in Page $8$, Section $2$, Example $3$.
By letting $n=m$ and $j=i$, we can rewrite this lemma and obtain the following: $$\begin{align}\sum_{i=1}^m(2i-1)&=(2\times 1-1)+\sum_{i=2}^m(2i-1) \\ \Leftrightarrow m^2&= 1+\sum_{i=2}^m(2i-1) \\ \Leftrightarrow \sum_{i=2}^m(2i-1)&= m^2-1 \\ \Leftrightarrow 2S&=m^2-1\end{align}$$ $$\boxed{\therefore S=\frac{m^2-1}{2}}\tag*{$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \ \ \ \; \ \,$}$$
After rewriting and simplifying the $x$ coordinate in the ordered pair of the conjecture, it follows that it is equivalent to $$\bigg(\pm\frac{m^2-1}{2m}, \frac{m^2-1}{2}\bigg).$$ And now... where do I go? Factorising, where $m^2-1=(m+1)(m-1)$, does not appear to help. Additionally, I split the $x$ coordinate to write $$\bigg(\pm\bigg(\frac{m^2-4}{2m}+\frac{3}{2m}\bigg),\frac{m^2-1}{2}\bigg)$$ knowing that $$\frac{m^2-4}{2m}=\frac{m^2-2^2}{2m}=\frac{m}2+\frac{2}{m}=\frac{m}2-\bigg(-\frac{2}m\bigg).\tag{$\ast\ast$}$$ Now if we let $g_1$ and $g_2$ be gradients such that $g_1\perp g_2$ and $g_1=m\div 2$ then Eq. $(\ast\ast)$ will be equivalent to $g_1-g_2$. But this did not help either, albeit exploring in the realm of gradients.

May anyone like to share some input? Did I take the right approach, or is this tedious and not very efficient? If you'd like, you can either tell me where to go from here, or draft up a new proof (though I hope the latter does not take up too much of your time). And, since we are dealing with graphs, you can perhaps use the Desmos Graphing Calculator (or, if you like to live on the dangerous side, you can try answering this with what I used — a pen, ruler and graphing paper!). I will accept any kind of answer, even if it may be a mere hint, as long as the hint serves useful.
With that being said, my general skill level is strongest probably up to Year $12$ mathematics... I believe... I mean, I am in Year $10$ and most of my maths that I know — of which is completely foreign to the Year $10$ maths curriculum at school — is self-taught (thanks, MSE!). After taking some Year $11$ Maths Methods classes, specifically on algebra, I found it very easy, so in order to keep answers understandable (and to not waste your time), I will safely say my skill level does not reach university standards, although it might...?
Thank you in advance!
$\boxed{\stackrel{\centerdot\,\centerdot}{\smile}}$
 A: The conjecture actually never holds (no wonder why you can't prove it!). For example, when $m=5$, $S=17/2$. Now the equation $y=5x$ meets the line $y=(5^2+1)/2=13$ at $(x,y)=(13/5,13)$, whose $y$-coordinate is not the same as $S$. 

Using the identity that $\sum_{i=1}^n i=n(n+1)/2$, \begin{align}S&=1+\sum_{i=2}^{m-1}\left(i-\frac12\right)=1-\left(1-\frac12\right)+\sum_{i=1}^{m-1}\left(i-\frac12\right)\\&=\frac12+\sum_{i=1}^{m-1}i-\sum_{i=1}^{m-1}\frac12=\frac12+\frac{(m-1)m}2-\frac{m-1}2\\\implies S&=\frac12(1+m^2-m-(m-1))=\frac{m^2-2m+2}2.\end{align} The intersection point between $y=\pm mx$ and $y=\frac{m^2+1}2$ occurs at $x=\pm\frac{m^2+1}{2m}$. However, $$\left(\pm\frac Sm,S\right)=\left(\pm\frac{m^2-2m+2}{2m},\frac{m^2-2m+2}2\right)$$ which is different to $$\left(\pm\frac{m^2+1}{2m},\frac{m^2+1}2\right)$$ As an addendum, for them to be the same, one can equate coordinates to get $$\frac{m^2-2m+2}{2m}=\frac{m^2+1}{2m}\implies m^2-2m+2=m^2+1\implies 2m=1\implies m=\frac12$$ which is not possible as $m\in\Bbb Z^+$.
