A Diophantine Equation Related to the Markoff Numbers

Question

Consider the following Diophantine equation. $$5(p^2+q^2+r^2+s^2+t^2)^2-7(p^4+q^4+r^4+s^4+t^4)=90pqrst$$ This equation was discussed at The Diophantine equation $5(p^2+q^2+r^2+s^2+t^2)^2 - 7(p^4+q^4+r^4+s^4+t^4) = 90pqrst$

The author was interested about the source, not the solution. I found the problem quite interesting. I wrote a code which generated all the solutions $1 \leq p \leq q \leq r \leq s \leq t \leq 500$.

p q r s t
1 1 1 1 1
1 1 1 1 2
1 1 1 2 2
1 1 1 2 4
1 1 2 3 3
1 1 2 3 6
1 2 4 7 7
1 2 7 7 17
1 2 34 47 127
1 17 42 63 124
2 7 19 47 59
2 13 39 97 99
3 3 6 42 87
3 3 27 87 189
6 21 138 417 483
11 11 11 154 286

According to the magazine mentioned in the link, this equation has a relation to the Markov numbers (https://en.wikipedia.org/wiki/Markov_number).

I wonder if there are similar tree like structure for the solutions of this problem.

Answer 1

The answer depends on the precise meaning of your "similar tree like structure". Consider a very general situation. Fix a symmetric polynomial $\,P(x_1,\dots,x_n).\,$ Define $$ T:=\{(x_1,\dots,x_n)\in\mathbb{Z}^n:P(x_1,\dots,x_n)=0\}.\tag{1}$$ Define a graph $\,G\,$ where the vertices $\,V\,$ are equivalence classes of $\,T\,$ where two elements of $\,T\,$ are regarded as equivalent if they are permutations of each other. Define two vertices as adjacent if they are equal up to permutation except for at most one component. Because $\,P\,$ is a polynomial, there can be at most $\,n-1\,$ vertices adjacent to a given vertex. If there are no cycles, then the graph is a tree. In the case of the Markov numbers, $\,P(x,y,z) = x^2+y^2+z^2-3xyz.\,$ In the case in your question, $$ P(p,q,r,s,t)=5(p^2+q^2+r^2+s^2+t^2)^2-7(p^4+q^4+r^4+s^4+t^4)-90pqrst. \tag{2}$$ It is probably true that there are no cycles but I have no proof for that, even if the numbers are restricted to positive integers. Thus, the solutions form a tree like structure.