I considered this problem, but, I cannot solve.
$a,b,c$ are a real numbers with $(a,b,c)\neq (0,0,0)$.
$k$ is defined as follows. $$k=\frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{(a^5+b^5+c^5)}$$ if $ab+bc+ca=0$, How many possible values of $k$ are there?
Firstly, I substitute $c=-\frac{ab}{a+b}$ into the equation of $k$, when the value of a+b is not $0$. Then,I got this equation.
$$k=1-\frac{2 a^8 b^2 + 8 a^7 b^3 + 16 a^6 b^4 + 20 a^5 b^5 + 16 a^4 b^6 + 8 a^3 b^7 + 2 a^2 b^8}{a^{10} + 5 a^9 b + 10 a^8 b^2 + 10 a^7 b^3 + 5 a^6 b^4 + a^5 b^5 + 5 a^4 b^6 + 10 a^3 b^7 + 10 a^2 b^8 + 5 a b^9 + b^{10}}=$$
However, this equation is too messy.
Secondly,I made an elementary symmetric polynomial from the equation of $k$. Then, I got this equation. $$k=\frac{(a+b+c)^3+3abc}{(a+b+c)^3+5abc}$$ However, I cannot proceed from here.
If you can solve this, could you tell me the answer or the hints? I'm sorry my broken English,I'm Japanese.
Looking at integers only. One infinite family is $$ a = u^2 - v^2 \; , \; \; \; b = 2(uv+v^2) \; , \; \; c = 2(-uv+v^2) \; , \; \; $$
Your ratio, fifth powers in the denominator, comes out $$ \frac{u^6 - 3 u^4 v^2 + 51 u^2 v^2 + 15 v^6}{u^6 - 11 u^4 v^2 + 67 u^2 v^4 + 7 v^6} $$
This is homogeneous, so we may divide through by $v^6,$ then draw a grph using $x = \frac{u}{v}.$ allowing for irrational $x$ gives us all values from $1$ to $15/7.$ Out of these there are infinitely many rational values when $u,v$ go back to being integers
Some output, integer pairs $(u,v)$ that are coprime with $u+v$ odd.