Another Diophantine equation

We can divide this problem in 3 cases.

Case I: $p = 0$ . Then $12q^4 = 0$ , and so $q = 0$ .

Case II: $q = 0$ . Then $5p^4 = 0$ , and so $p = 0$ , which is the same solution from case I. The trivial solution $(0, 0)$ works here.

Case III: $p*q \not= 0$ . Take the original equation: $5p^{4} + 12q^{4} = 6pq^{3} + 3p^{2}q^{2} + 9p^{3}q$

Suppose that $(p, q)$ satisfies the original equation. Notice that if we multiply this solution by any non-zero real number $n$ , to obtain a solution of the type $(np, nq)$ , it is also a solution of the equation, since:

$5(np)^{4} + 12(nq)^{4} = 6(np)(nq)^{3} + 3(np)^{2}(nq)^{2} + 9(np)^{3}(nq)$ Dividing this by $n^{4}$ gives us the original equation.

Thus, if $(p, q)$ is a non-trivial solution and $n$ a non-zero real number, then $(np, nq)$ is a solution of the equation as well, so we'd have an infinite number of solutions besides the trivial one.

Now, we shall prove that there are no solutions other than the trivial.

Dividing the original equation by $q^{4}$ yields: $5(p/q)^{4} + 12 = 6(p/q) + 3(p/q)^{2} + 9(p/q)^{3}$ . If $p/q = r$ , then:

$5r^{4} + 12 = 6r + 3r^{2} + 9r^{3}$ , thus $5r^{4} - 9r^{3} - 3r^{2} - 6r + 12 = 0$ . According to the rational root theorem, if this equation has a solution of the form $a/b$ such that $a$ and $b$ are coprime, then $a$ must divide 12 and $b$ must divide 5. We have the candidate solutions: $(1/5, 2/5, 3/5, 4/5, 1, 6/5, 2, 12/5, 3, 4, 6, 12)$ and their negative counterparts. It's not hard to verify that none of these solutions work, though it is a time-consuming process. Thus, $r$ cannot be a rational number, so $p/q$ cannot be a rational number and then there are no solutions for which $p$ and $q$ are non-zero integers. Thus, there is only one valid solution, the trivial one.