Question About the Logic of my Proof Okay, I am working on the following relatively simple problem:

Let $f(x) = |x-3| + |x-1|$ for all $x \in \mathbb{R}$. Find all $t$ for which $f(t+2) = f(t)$.

So, if $f(t+2)=f(t)$, the is equivalent to $|t+1| = |t-3|$. Thus, if this holds, one can square both sides and arrive at $t=1$. So, this value of $t$ is a necessary condition, but prima facie it isn't the only value. To show sufficiency, could I let $t = 1 + \epsilon$, plug it into the above equation, deduce that $\epsilon = 0$, and conclude that $t=1$ is the only value? Would that go into showing this is the only value?

When squaring both sides of an equation you can't lose solutions you could only get extra false solutions. Since $t=1$ satisfies $|t+1|=|t-3|$ that means $t=1$ is not a false solution and is the only solution.
