3. Find all functions $g: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$, such that for all $m, n \in \mathbf{N}_{+}, (g(m) + n)(m + g(n))$ is a perfect square.
(US Proposal)
3. $g(n)=n+c$, where the constant $c$ is a non-negative integer.

First, the function $g(n)=n+c$ satisfies the problem (because at this time
$$
(g(m)+n)(m+g(n))=(n+m+c)^{2}
$$

is a perfect square).
First, we prove a lemma.
Lemma If a prime $p$ divides $g(k)-g(l)$, where $k$ and $l$ are positive integers, then $p \mid(k-l)$.

Proof In fact, if $p^{2} \mid (g(k)-g(l))$, let $g(l)=g(k)+p^{2} a$, where $a$ is some integer.

Take an integer $d>\max \{g(k), g(l)\}$, and $d$ cannot be divided by $p$.
Let $n=p d-g(k)$. Then
$$
\begin{array}{l}
n+g(k)=p d . \\
\text { Hence } n+g(l)=p d+(g(l)-g(k)) \\
=p(d+p a)
\end{array}
$$

can be divided by $p$, but not by $p^{2}$.
By the problem's condition, $(g(k)+n)(g(n)+k)$ and $(g(l)+n)(g(n)+l)$ are both perfect squares. Therefore, if they can be divided by the prime $p$, they can also be divided by $p^{2}$. Thus,
$$
p|(g(n)+k), p|(g(n)+l) .
$$

Hence $p \mid [(g(n)+k)-(g(n)+l)]$,
i.e., $p \mid(k-l)$.
If $p \mid (g(k)-g(l))$, take the same integer $d$, let $n=p^{3} d-g(k)$. Then the positive integer
$$
g(k)+n=p^{3} d
$$

can be divided by $p^{3}$, but not by $p^{4}$, the positive integer
$$
g(l)+n=p^{3} d+(g(l)-g(k))
$$

can be divided by $p$, but not by $p^{2}$. Thus, by the problem's condition,
$$
p|(g(n)+k), p|(g(n)+l) .
$$

Hence $p \mid [(g(n)+k)-(g(n)+l)]$,
i.e., $p \mid (k-l)$.
Back to the original problem.
If there exist positive integers $k$ and $l$ such that $g(k)=g(l)$, then by the lemma, $k-l$ can be divided by any prime $p$.
Thus, $k-l=0$, i.e., $k=l$. Therefore, $g$ is injective.
Consider the numbers $g(k)$ and $g(k+1)$.
Since $(k+1)-k=1$, by the lemma, $g(k+1)-g(k)$ cannot be divided by any prime.
Hence $|g(k+1)-g(k)|=1$.
Let $g(2)-g(1)=q(|q|=1)$. Then by mathematical induction, it is easy to know
$$
g(n)=g(1)+(n-1) q .
$$

If $q=-1$, then for $n \geqslant g(1)+1$, we have $g(n) \leqslant$ 0, which is a contradiction. Therefore, $q=1$.

Hence $g(n)=n+(g(1)-1)$ for all $n \in \mathbf{N}$, where $g(1)-1 \geqslant 0$.

Let $g(1)-1=c$ (constant). Hence $g(n)=n+c$, where the constant $c$ is a non-negative integer.