Here it is. Let $a_n=(a_1,m)$ and $b_n=(b_1,k)$ be the first term, and common differences of both sequences. Using the identity, $(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2$, we arrive at, $$ (a_n^2+a_{n+1}^2)(b_n^2+b_{n+1}^2)=(a_nb_n + a_{n+1}b_{n+1})^2 + (a_nb_{n+1}-a_{n+1}b_n)^2; $$and, $$ (a_n^2+b_n^2)(a_{n+1}^2+b_{n+1}^2)=(a_na_{n+1}+b_nb_{n+1})^2+(a_nb_{n+1}-b_na_{n+1})^2. $$Now, we concentrate on the common second terms above. Noting that $b_{n+1}=b_1+nk$ and $a_{n+1}=a_1+mk$, $$ a_nb_{n+1}-a_{n+1}b_n = a_n(b_n+k)-b_n(a_n+m)=a_nk-b_nm =(a_1+(n-1)m)k-(b_1+(n-1)k)m=a_1k-b_1m, $$which is a constant. Lemma$(a_n^2+a_{n+1}^2)(b_n^2+b_{n+1}^2)$ is a perfect square, for only finitely many values of $n$; if $D\neq 0$. Proof. Assume the claim is false; let it be a perfect square infinitely many times. Passing to a subsequence if necessary, we let $u_n^2$ to be this sequence, and write $u_n^2 = D^2+v_n^2$ with $v_n =|a_nb_{n+1}-a_{n+1}b_n|$. Observe that, $u_n>|v_n|$; hence, $$ D^2+v_n^2 \geq (v_n+1)^2 = v_n^2+2v_n+1\implies \frac{D^2-1}{2}\geq v_n. $$However, since $u_n\to\infty$ as $n$ grows; and $D$ is fixed, we must have that $v_n$ grows, too; which is a contradiction. Similar to above, one can immediately modify the previous proof, and obtain that $(a_n^2+b_n^2)(a_{n+1}^2+b_{n+1}^2)$ is a perfect square, for only finitely many values of $n$, if $D\neq 0$. Hence, we must have that $D=0$. With this, $$ a_1k = b_1 m \implies m\mid a_1 k \implies m\mid k, $$since $(a_1,m)=1$. Similarly, we also have $k\mid b_1m\implies k\mid m$; hence, $k=m$ must hold. From here, we get $a_1=b_1$; and therefore, both sequences are the same.