I have the following problem.
Let $\alpha,\beta:I\rightarrow\mathbb{R}^3$ 2-regular curves (curvature $k\not=0$ for all $t\in I$), arc-length parametrized. Let $(t_\alpha, n_\alpha, b_\alpha)$ and $(t_\beta, n_\beta, b_\beta)$ be the Frenet frames of $\alpha$ and $\beta$, respectively. Suppose we have $$t_\alpha(u)+b_\alpha(u)=t_\beta(u)+b_\beta(u) \quad \text{for all } u\in I$$ and suppose that $k_\alpha(u)\not=\tau_\alpha(u)$ for all $u\in I$, where $k$ represents the curvature and $\tau$ the torsion of each curve. Prove that it exists a vector $v\in\mathbb{R}^3$ such that $\beta(u)=v+\alpha(u)$ for all $u\in I$.
My idea was to show that $k_\alpha=k_\beta$ and $\tau_\alpha=\tau_\beta$ in order to use Fundamental theorem of curves, which would prove the statement.
I used the given equality $t_\alpha+b_\alpha=t_\beta+b_\beta$ and the Frenet-Serret formulas to show that $|n_\alpha| = |n_\beta| = 1$ and have the same direction. This gives us two cases:
For $n_\alpha=n_\beta$, I could check that $k_\alpha=k_\beta$ and $\tau_\alpha=\tau_\beta$, thus we get to the solution.
However, I'm struggling to solve the case $n_\alpha=-n_\beta$. I would really appreciate help with this.