How to prove uniqueness of |a - c| = |b - c| So I'm working on a problem, and the problem asks to prove that |a - c| = |b - c| has one unique integer solution for any odd integers a and b.
I have proven that there exists a number (a + b)/2 which satisfies this property for all odd integers a and b.
My problem is that proving that the average fits this property does not satisfy the conditions that it a) must be an integer and b) must be unique. I can show this intuitively but not mathematically. Any help would be appreciated.
 A: In general, the simplest way to show that there is a unique $\;c\;$ such that $\;P\;$, is to rewrite $\;P\;$ to an equivalent expression of the form $\;c = \ldots\;$.  (On the other hand, if you only have to prove that there exists some $\;c\;$, then it is sufficient to prove just the $\;P \;\Leftarrow\; c = \ldots\;$ direction.)
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\followsfrom}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$So our next step is to try and solve $$\tag{0} |a-c| = |b-c|$$ for $\;c\;$, where $\;c\;$ is an integer, for example by the following calculation:
$$\calc
|a-c| = |b-c|
\calcop\equiv{basic property of $\;|\cdot|\;$}
a-c = b-c \;\lor\; -(a-c) = b-c
\calcop\equiv{simplify left hand side; simplify right hand side}
a = b \;\lor\; 2 \times c = a+b
\calcop\equiv{divide by 2}
a = b \;\lor\; c = \tfrac{a+b}2
\endcalc$$
So we have not succeeded in proving a statement of the form $\;|a-c| = |b-c| \;\equiv\; c = \ldots\;$, since that $\;a=b \;\lor\;\;$ part is still there.
However, our work was not in vain, since this gives us a counterexample: if $\;a=b\;$, then any integer $\;c\;$ is a solution for $\ref 0$.  And since there is more than one integer, there is no unique solution to $\ref 0$.
So, to give a specific counterexample, for $\;a = b = 0\;$ every integer $\;c\;$ satisfies $\ref 0$.
(On the other hand, if you only have to prove that there exists some $\;c\;$, then you can simply continue the above calculation
$$\calc
\ldots
\calcop\followsfrom{logic: weaken}
c = \tfrac{a+b}2
\endcalc$$
Now as you correctly observed in a comment, $\;\tfrac{a+b}2\;$ is an integerm since $\;a+b\;$ is even, since $\;a,b\;$ are both odd.  So you've proven something of the form $\;|a-c| = |b-c| \;\followsfrom\; c = \ldots\;$, so you've proven existence.)
