This question is inspired by this one. It comes in two parts.
Question 1. Determine all positive integers $k$ such that there are positive integers $a$, $b$, and $c$ such that
$$\frac{a^2+b^2+c^2}{bc+ca+ab}=k\,.\tag{*}$$
Question 2. For each positive integer $k$ discovered in Question 1, what are all triples $(a,b,c)$ of positive integers such that the condition (*) is satisfied?
Here are three values of $k$ that have the required property.
-
Case I: $k=1$. All solutions $(a,b,c)$ are of the form $$(a,b,c)=(n,n,n)$$ where $n$ is a positive integer.
-
Case II: $k=2$. It can be proven by Vieta jumping that each solution $(a,b,c)$ is a permutation of
$$\big(tm^2,tn^2,t(m+n)^2\big)\tag{#}$$
for some positive integers $t$, $m$, and $n$ (we can assume that $m$ and $n$ are relatively prime). A proof of this claim can be seen in the hidden portion below. -
Case III: $k=5$. All solutions can be found in this link.
Are there other values of $k$ with the required property? If so, are there infinitely many of them?
Here is a proof sketch for my claim when $k=2$ if you would like to read. Let $S$ denote the set of solutions $(a,b,c)\in\mathbb{Z}_{>0}^3$ to (*). Define a similarity relation $\sim$ on $S$ which is an equivalence relation on $S$ generated by requiring that each triple $(a,b,c)\in S$ is similar to any permutation of $(a,b,c)$, and that $(a,b,c)$ is similar to $(a,b,2a+2b-c)$, provided that $(a,b,2a+2b-c)$ is also in $S$. Pick an equivalence class $C$ of $S$ induced by $\sim$, and suppose that $(a,b,c)$ is its minimal triple in the sense that $a+b+c$ is the smallest among all triples in $C$ that is not of the form (#). We may assume without loss of generality that $a\leq b\leq c$. Note that either $2a+2b-c\leq 0$ or $(a,b,2a+2b-c)$ is a "smaller" triple than $(a,b,c)$ in $C$ that is not of the form (#). Show that $c=2a+2b$ must holds, and this implies $b=c$. It then follows that $(a,b,c)=(t,t,4t)=\big(1^2t,1^2t,(1+1)^2t\big)$ for some positive integer $t$, and this is a contradiction.
Best Answer
There is such a solution if and only if both $k-1$ and $k+2$ have (well, different) integer expressions as some $u^2 + 3 v^2.$
The justification for that is in several answers I posted at
Find a solution: $3(x^2+y^2+z^2)=10(xy+yz+zx)$
$$ $$ $$ $$
Given $$ p^2 + 3 q^2 = 2 + k, $$ $$ r^2 + 3 s^2 = 4(k-1), $$ we can solve $$ (x^2 + y^2 + z^2) = k (yz + zx + xy) $$ with $$ x = 2 p^2 + 6 q^2 - p r - 3 p s + 3 q r - 3 q s, $$ $$ y = 2 p^2 + 6 q^2 - p r + 3 p s - 3 q r - 3 q s, $$ $$ z = 2 p^2 + 6 q^2 + 2 p r + 6 q s. $$
I did not immediately realize, the process of Vieta Jumping lets us take a mixed solution and create one with all the same $\pm$ sign. Suppose $x < 0,$ $y > 0,$ $z>0.$ We do a single jump: $$ x \mapsto k(y+z) - x, $$ where the new $x$ value is then positive!
The permissible values of your $k$ from 2 to 1000 are
These all lead to solutions $(a,b,c) $ where it may be that some variables are negative, some positive.
Let me work up some of the smallest such $k,$ see whether positive solutions appear.
$$ k = 17; \; \; \; (377,17,5) $$
$$ k = 26; \; \; \; (418,13,3) $$
$$ k = 29; \; \; \; (1109,11,27) $$
BY RECIPE .........................................