Determine all positive integers $n$ with the following property: for every triple $(a, b, c)$ of positive real numbers, there is a triple $(k, \ell, m)$ of non-negative integers such that $a n^{k}, b n^{\ell}$ and $c n^{m}$ are the lengths of the sides of a (non-degenerate) triangle.
It is clear that $n=1$ does not work, because not every three positive real numbers $a, b,$ and $c$ are the lengths of a triangle. We now first prove that $n \geq 5$ does not work by considering the triplet $(1,2,3)$. Suppose there exist $k, \ell, m$ such that $n^{k}, 2 n^{\ell},$ and $3 n^{m}$ are the lengths of the sides of a triangle. First, note that no two of these three numbers can be equal, since $n \neq 2,3$. By possibly scaling the triangle by a factor of $n$ several times, we can assume that one of the numbers $k, \ell,$ and $m$ is equal to 0. Suppose the other two are both positive, then the corresponding side lengths are both multiples of $n$. Their difference is thus at least $n$, while the third side is at most 3. This contradicts the triangle inequality. Therefore, at least two of $k, \ell,$ and $m$ must be equal to 0. The corresponding two side lengths have a sum of at most 5, so the third side must be less than 5. From $n \geq 5$ it follows that this third side also does not have a factor of $n$ in its length. Thus, $k, \ell,$ and $m$ are all equal to 0, but then 1, 2, and 3 would have to be the lengths of the sides of a triangle, while $3=2+1$. Contradiction. We conclude that $n \geq 5$ does not work.

Now consider $n=2,3,4$. We construct $(k, \ell, m)$ as follows. Take a triplet $(a, b, c)$. If these are already the sides of a triangle, we take $k=\ell=m=0$. Otherwise, there is a triangle inequality that does not hold, say without loss of generality that $a \geq b+c$. Now multiply the smallest of $b$ and $c$ by $n$. If this does not make the right-hand side greater than $a$, multiply the smallest (of the new two terms) by $n$ again. Generally, if $a \geq n^{i} b+n^{j} c$, then multiply the smallest of $n^{i} b$ and $n^{j} c$ by $n$ and check again if the inequality holds. This process is guaranteed to stop, because there is an $i$ such that $n^{i}>a$. Now consider the $i$ and $j$ such that $a \geq n^{i} b+n^{j} c$ and this is the last step where this inequality holds. Without loss of generality, assume that $n^{i} b \leq n^{j} c$, then $a < n^{i+1} b + n^{j} c$. We claim that we can take $(k, \ell, m)$ to be one of $(0, i+1, j)$ and $(0, i+1, j+1)$.

By definition, $a < n^{i+1} b + n^{j} c$ and further, $n^{j} c < n^{i} b + n^{j} c \leq a < a + n^{i+1} b$, so if $(0, i+1, j)$ does not work, it must be because $n^{i+1} b \geq a + n^{j} c$. Furthermore, $n^{i} b \leq n^{j} c$, so $n^{i+1} b \leq n^{j+1} c < a + n^{j+1} c$. Also, of course, $a < n^{i+1} b + n^{j+1} c$, so if $(0, i+1, j+1)$ does not work, it must be because $n^{j+1} c \geq a + n^{i+1} b$. We will derive a contradiction in the case that both triplets do not work. We then know that both $n^{i+1} b \geq a + n^{j} c$ and $n^{j+1} c \geq a + n^{i+1} b$.

Adding these inequalities and canceling $n^{i+1} b$ on both sides gives $n^{j+1} c \geq 2a + n^{j} c$, or $(n-1) n^{j} c \geq 2a$. Thus,

$$
n^{i+1} b \geq a + n^{j} c \geq a + \frac{2a}{n-1} = \left(1 + \frac{2}{n-1}\right) a
$$

which implies

$$
a \geq n^{i} b + n^{j} c \geq \frac{1}{n} \left(1 + \frac{2}{n-1}\right) a + \frac{2}{n-1} a = \frac{(n-1) + 2 + 2n}{n(n-1)} a = \frac{3n + 1}{n(n-1)} a.
$$

Thus, $3n + 1 \leq n(n-1)$. For $n=2,3,4$, this gives $7 \leq 2$, $10 \leq 6$, and $13 \leq 12$, respectively, which are contradictions in all cases.

We conclude that the values of $n$ that work are exactly $n=2,3,4$.

Alternative solution for the first part. We prove in a different way that $n \geq 5$ does not work. We consider the triplet $(1,2,3)$ again. Suppose there exist $k, \ell, m$ such that $n^{k}, 2 n^{\ell},$ and $3 n^{m}$ are the lengths of the sides of a triangle. From the triangle inequality, we have

$$
\begin{aligned}
n^{k} & < 2 n^{\ell} + 3 n^{m} \\
2 n^{\ell} & < n^{k} + 3 n^{m} \\
3 n^{m} & < n^{k} + 2 n^{\ell}
\end{aligned}
$$

If $m \leq \ell-1$ and $k \leq \ell$, then

$$
n^{k} + 3 n^{m} \leq n^{\ell} + \frac{3}{n} n^{\ell} = \left(1 + \frac{3}{n}\right) \cdot n^{\ell} \leq 2 n^{\ell}
$$

which contradicts the second triangle inequality. Therefore, $m \geq \ell$ or $k \geq \ell+1$. Similarly, from the first triangle inequality, we can deduce that $\ell \geq k$ or $m \geq k$, and from the third that $k \geq m+1$ or $\ell \geq m+1$.

We now distinguish two cases: $m \geq \ell$ or $k \geq \ell+1$. Suppose first that $m \geq \ell$. Then $\ell \geq m+1$ does not hold, so $k \geq m+1$ must hold. But then $m \geq k$ does not hold, so $\ell \geq k$ must hold. We find $m \geq \ell \geq k \geq m+1$, a contradiction. Suppose next that $k \geq \ell+1$. Then $\ell \geq k$ does not hold, so $m \geq k$ must hold. But then $k \geq m+1$ does not hold, so $\ell \geq m+1$ must hold. We find $k \geq \ell+1 \geq m+2 \geq k+2$, a contradiction. We conclude that $n \geq 5$ does not work.