Let $s$ and $t$ be two relatively prime odd numbers. We want to know the $d=\gcd(s-t,s+t).$ Let $d=2^lm$ thus $s-t=2^lmk_1$ and $s+t=2^lmk_2$ for $k_1$ and $k_2$ odd numbers and $\gcd(k_1, k_2)=1.$ Summing and subtracting both sides of two equality-es and diving by $2$ results in $s=2^lm \frac{(k_1+k_2)}{2}$ and $t=2^lm \frac{(k_2-k_1)}{2}$ in which $\frac{(k_2 \pm k_1)}{2}$ are integers as both $k_1$ and $k_2$ are odd numbers and because $s$ and $t$ are relatively prime implies $l=0$ which is not possible since both $s-t$ and $s+t$ are even so must $l \ge 1$ !! Contradiction?
A rewritten version of this same question, for the author's benefit:
Let $s$ and $t$ be two distinct relatively prime odd numbers.
We want to compute $d=\gcd(s-t,s+t).$ Note that $s-t$ and $s+t$ are both nonzero because $s$ and $t$ are distinct.
Let $d=2^\ell m$, where $m$ is odd. Then since $s-t$ and $s+t$ are both multiples of $d$, we have
\begin{align}
s-t&=2^\ell mk_1 \text{and}\\
s+t&=2^\ell mk_2
\end{align}
where $k_1$ and $k_2$ are both odd. Furthermore, $\gcd(k_1, k_2)=1,$ for if it were some other number $u > 1$, then $ud$ would divide both $s+t$ and $s-t$, contradicting the fact that $d$ is their greatest common divisor.
Finally, since both $s-t$ and $s+t$ are even and nonzero, we must have $\ell \ge 1$, which we'll use shortly.
Summing and subtracting both sides of these two equalities and dividing by $2$ gives
\begin{align}
s&=2^\ell m \frac{(k_1+k_2)}{2} \text{ and}\\
t& =2^\ell m \frac{(k_2-k_1)}{2}
\end{align}
Note that because $k_1$ and $k_2$ are both odd, their sum and difference are both even, hence
$\frac{(k_2 \pm k_1)}{2}$ are both integers.
Now $s$ and $t$ are relatively prime (given), so $\ell$ must be zero (otherwise $2^\ell$ would divide both, and the gcd could not be $1$).
On the other hand, we showed above that $\ell \ge 1$. That's a contradiction.
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Now there's definitely something screwy in the "proof" above, because Jyrki's example shows that the gcd is actually $2$. But you'll have a lot easier time finding the error now that the "proof" is actually written clearly and coherently.
Best Answer
$k_1$ and $k_2$ don't have to be odd numbers, only one of them has to be. If $s$ and $t$ are both odd, $s+t$ and $s-t$ are both even, so you know that $d$ is also even.