$a_{m^2}=a_m^2,a_{m^2+k^2}=a_ma_k$ sequence 
Sequence $\{a_n\},n\in\mathbb N_+$ with all terms positive integers satisfy $a_{m^2}=a_m^2,a_{m^2+k^2}=a_ma_k$. Find $\{a_n\}$.

I suppose all terms of $\{a_n\}$ are $1$. This problem makes me think of a lot of conclusions, including

*

*$n\in\mathbb{Z}_+$ can be written as the sum of two squares as long as for every prime $p\equiv3\pmod4$ there's $2\mid V_p(n)$.

*$(m^2+n^2)^2=(m^2-n^2)^2+(2mn)^2$.

*$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2=(ac-bd)^2+(ad+bc)^2$.

Perhaps we can let the first non-$1$ term of the sequence be $a_s$ and derive a contradiction?

We claim that all $a_{n}$ are $1$, we will use proof by contradiction to make our claim clear.
Suppose $\{ a_{n}\}$ is some sequence which is not the trivial sequence composed of $1$'s and satisfies your requirements.
$\textbf{Let}$ $v$ $\textbf{be the smallest positive integer so that}$ $a_{v} \neq 1$. Note that $a_{1} = 1$ as $a_{1} = a_{1}^2$ and $a_{1} \geq 1$, and $a_{2} = a_{1}^2 = 1.$ Thus $v \geq 3$. Note that $v$ is $\textbf{not}$ a sum of two positive squares, because if we have $v = c^2 +d^2$ for $c,d \in \mathbb{N}_{+}$ then $a_{v} = a_{c}a_{d} = 1$. $\textbf{Nor}$ is $v$ a square as if $v = j^2$ then $a_{v} = a_{j} = 1$.
$\textbf{Lemma:}$ For each $w \in \mathbb{N}$, there exists $\alpha_{w} \in \{1,3,5\}$ so that $\frac{w-5\alpha_{w}}{3}$ and $\frac{w+4\alpha_{w}}{3}$ are integers and
$w^2+(\frac{w-5\alpha_{w}}{3})^2 = (w-\alpha_{w})^2+(\frac{w+4\alpha_{w}}{3})^2$
In the above lemma note that when $w \geq 10$ we have $|\frac{w-5\alpha_{w}}{3}|\leq w$, $0 < (w-\alpha_{w}) < w,$ and $\frac{w+4\alpha_{w}}{3}$.
Due to the above lemma, if $v \geq 11$ then there are two cases;
$\textbf{Case 1:}$ $\frac{v-5\alpha_{v}}{3} = 0$,
In this case one has that $v^2 = (v-\alpha_{v})^2+(\frac{v+4\alpha_{v}}{3})^2$.
Thus $$a_{v^2}  = a_{v-\alpha_{v}}a_{\frac{v+4\alpha_{v}}{3}} = 1,$$ but $a_{v^2} = a_{v} > 1$ which is impossible. Hence this case is impossible.
$\textbf{Case 2:}$ $\frac{v-5\alpha_{v}}{3} \neq 0$.
In this case, note that $$a_{v^2+(\frac{v-5\alpha_{v}}{3})^2} = a_{v}a_{\frac{v-5\alpha_{v}}{3}} = a_{v} > 1$$
but also
$$a_{v^2+(\frac{v-5\alpha_{v}}{3})^2} = a_{(v-\alpha_{v})^2+(\frac{v+4\alpha_{v}}{3})^2} = 1$$
which is impossible. Thus this case is impossible.
Hence we know that $3 \leq v \leq 10$. Now it is time for brute force...
$v \neq 3$ as $a_{9+16} = a_{3}a_{4} = a_{3}a_{2} = a_{3}$ and $a_{25} = a_{5} = a_{1+4} = a_{1}a_{2} = 1$ thus $a_{3} = 1$
$v \neq 4$ as $4$ is a square (see above)
$v \neq 5$ as $5 = 1+2^2$ (see above)
$v \neq 6$. To see this note that $6^2 +8^2 = 100$. Thus $a_{100} = a_{6}a_{8}=a_{6}(a_{2})^2 = a_{6}$ and $a_{100} = a_{10} = a_{9+1} = a_{3}a_{1} = 1$. Thus $a_{6} = a_{1}$.
$v \neq 7$. To see this note that $7^2+1^2 = 50 = 5^2+5^2$. Thus $$1 = a_{5}^2 = a_{50} = a_{7}$$
$v \neq 8$ as $8 = 4+4$
$v \neq 9$ as $9$ is a square.
$v \neq 10$ as $10 = 9+1$.
$\textbf{Thus there is no smallest integer}$ $v$ $\textbf{such that}$ $a_{v} \neq 1$. This is a contradiction towards our assumption.
