5. Find all functions $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$, such that for all positive integers $a, b$, there exists a non-degenerate triangle with side lengths $a, f(b), f(b+f(a)-1)$ (a triangle is called non-degenerate if its three vertices are not collinear).
(France)
5. The function $f$ that satisfies the requirements can only be $f(n)=n\left(n \in \mathbf{N}_{+}\right)$.

From the conditions and the discreteness of integers, for any positive integers $a, b$, we have
$$
\begin{array}{l}
f(b)+f(b+f(a)-1)-1 \geqslant a, \\
f(b)+a-1 \geqslant f(b+f(a)-1), \\
f(b+f(a)-1)+a-1 \geqslant f(b) .
\end{array}
$$

By substituting $a=1$ in equations (2) and (3), we get
$$
f(b)=f(b+f(1)-1)
$$

for any $b \in \mathbf{N}_{+}$.
If $f(1) \neq 1$, then the above equation implies that $f$ is a periodic function. Combined with the domain of $f$, this means $f$ is bounded.

Choose a positive integer $M$ such that $M \geqslant f(n)$ for all positive integers $n$ (i.e., $M$ is an upper bound of $f$). Substituting $a=2M$ in equation (1) leads to a contradiction.
Therefore, $f(1)=1$.
Substituting $b=1$ in equations (1) and (2), we get
$$
f(f(n))=n\left(n \in \mathbf{N}_{+}\right).
$$

If there exists $t \in \mathbf{N}_{+}$ such that $f(t) < t$, let $M'$ be an upper bound of $f(n)$ for $1 \leq n \leq t-1$. Let $n_{0}$ be the unique positive integer satisfying
$$
1 \leqslant n_{0} \leqslant t-1, n_{0} \equiv n(\bmod t-1)
$$

Then,
$$
\begin{array}{l}
f(n) \leqslant f\left(n_{0}\right)+\frac{t-2}{t-1}\left(n-n_{0}\right) \\
\leqslant \frac{M'}{t-1}+\frac{(t-2) n}{t-1}
\end{array}
$$

If $f(n) > M'$, and $n_{1}$ is not equal to any of $f(1), f(2), \cdots, f\left(M'\right)$, then by $f\left(f\left(n_{1}\right)\right)=n_{1}$, we have $f\left(n_{1}\right) > M'$.
By the above conclusion,
$$
n_{1} > f\left(n_{1}\right) > f\left(f\left(n_{1}\right)\right) = n_{1},
$$

which is a contradiction. This implies that for any $t\left(t \in \mathbf{N}_{+}\right)$, we have $f(t) \geqslant t$, and thus,
$$
t = f(f(t)) \geqslant f(t) \geqslant t.
$$

All inequalities must be equalities, so $f(n) = n\left(n \in \mathbf{N}_{+}\right)$.
Upon verification, $f(n) = n\left(n \in \mathbf{N}_{+}\right)$ satisfies the requirements.
Therefore, the function $f$ is $f(n) = n\left(n \in \mathbf{N}_{+}\right)$.