Problem 1835

It is clear that $n = 1$ does not satisfy the property, as not every three positive real numbers $a, b$, and $c$ are the lengths of the sides of a triangle.

We first show that any $n \ge 5$ cannot satisfy the property by considering the triple $(a, b, c) = (1, 2, 3)$. Suppose that there exist $k, \ell, m$ such that $n^k, 2n^\ell, 3n^m$ are the lengths of the sides of a triangle. As $n \neq 2, 3$, no three of these are equal. By repeatedly removing common factors $n$ if they exist, we can and do assume that one of $k, \ell, m$ is equal to 0. Suppose that the other two of those three integers are positive, then their corresponding side lengths both are multiples of $n$. Their difference then is at least $n$, while the third side has length of at most 3. This contradicts the triangle inequality. Hence of $k, \ell, m$, at least two must be zero. The corresponding two side lengths then sum to at most 5, so the third side must have length less than 5. As $n \ge 5$, this third side cannot have a factor $n$ either, so $k, \ell, m$ are all equal to 0. However, then 1, 2, 3 should be the lengths of the sides of a triangle, while $3 = 2+1$. This is a contradiction. We deduce that $n \ge 5$ does not satisfy the property.

Now consider $n = 2, 3, 4$. We construct $(k, \ell, m)$ as follows. Take a triple $(a, b, c)$. If its entries already are the lengths of the sides of a triangle, we take $k = \ell = m = 0$. Otherwise there exists a triangle inequality that is not satisfied. Without loss of generality, we assume that $a \ge b + c$. Multiply the lesser of $b$ and $c$ with $n$. If the right hand side still isn't larger than $a$, then take the new summands, and again multiply the lesser of the two by $n$. So: if $a \ge n^i b + n^j c$, then we multiply the lesser of $n^i b$ and $n^j c$ by $n$, and repeat if the inequality still holds. This process always stops at some point, as there exists $i$ such that $n^i > a$.

Consider the $i$ and $j$ such that $a \ge n^i b + n^j c$, such that applying the process makes the right hand side larger than $a$. We assume without loss of generality that $n^i b \le n^j c$, so we have $a < n^{i+1} b + n^j c$. We claim that we can take $(k, \ell, m)$ to be equal to either $(0, i+1, j)$ or $(0, i+1, j+1)$.

By definition, we have $a < n^{i+1} b + n^j c$, and we have $n^j c < n^i b + n^j c \le a < a + n^{i+1} b$. Therefore, if $(0, i+1, j)$ doesn't satisfy the property, then we must have $n^{i+1} b \ge a + n^j c$. Moreover $n^i b \le n^j c$, so $n^{i+1} b \le n^{j+1} c < a + n^{j+1} c$. We also have $a < n^{i+1} b + n^{j+1} c$, so if $(0, i+1, j+1)$ does not satisfy the property, then we must have $n^{j+1} c \ge a + n^{i+1} b$. We will derive a contradiction in case both triples do not satisfy the property, i.e. in case both $n^{i+1} b \ge a + n^j c$ and $n^{j+1} c \ge a + n^{i+1} b$.

Adding these inequalities, and subtracting $n^{i+1} b$ on both sides, yields $n^{j+1} c \ge 2a + n^j c$, or equivalently $(n-1)n^j c \ge 2a$. Therefore
$$n^{i+1}b \ge a + n^j c \ge a + \frac{2a}{n-1} = \left(1 + \frac{2}{n-1}\right) a,$$
from which follows that
$$a \ge n^i b + n^j c \ge \frac{1}{n} \left(1 + \frac{2}{n-1}\right) a + \frac{2}{n-1} a = \frac{(n-1) + 2 + 2n}{n(n-1)} a = \frac{3n+1}{n(n-1)} a.$$
So $3n+1 \le n(n-1)$. For $n=2, 3, 4$, this inequality reads $7 \le 2, 10 \le 6$, and $13 \le 12$, all of which are false. This is a contradiction.

Hence $n=2, 3, 4$ are precisely the values which satisfy the property. $\square$