Integral domain $D$ where $a^{100}=b^{100}$ and $a^{121}=b^{121}$

Question

Let $D$ be an integral domain and $a,b \in D$ such that $a^{100}=b^{100}$ and $a^{121}=b^{121}$. Show that $a=b$.

Firstly, as $D$ is an integral domain, we have that if $a^{n}=0$ for some $n \in \mathbb{N}$, then $a=0$.

Now, I wrote that:

$$a^{121}=b^{121} \iff a^{100}a^{21}=b^{100}b^{21} \iff b^{100}(a^{21}-b^{21})=0 \implies b^{100}=0 \lor a^{21}=b^{21}$$

If $b^{100}=0$, then $a^{100}=0$ and $a=b=0$; if $a^{21}=b^{21}$, in a similar manner:

$$a^{100}=b^{100} \iff a^{21 \times 4}a^{16}=a^{21 \times 4}b^{16}=0 \implies a^{21 \times 4}=0 \lor a^{16}=b^{16}$$

$$a^{6 \times 16}a^{4}=b^{6 \times 16}b^{4}=0 \implies a^{6 \times 16}=0 \lor a^4=b^4$$

And finally,

$$a^{121}=b^{121} \iff a^{4 \times 30}a=a^{4 \times 30}b \implies a^{4 \times 30} \lor a=b$$

And we arrive at $a=b$. Is this argument correct? Can this be solved in a quicker way?