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.