Are there integer solutions of $m^4+m^2n^2+n^4=k^2$?

Question

The recent question about Sets of integers with same sum and same sum of reciprocals, with its wealth of solutions, raises naturally the question: Can we require all those numbers to be squares? Of course, no pigeonhole principle for that.

The family of solutions given in one of the answers would yield a start for a 3-term solution in squares if we can solve $$m^4+m^2n^2+n^4=k^2,$$ so this would be a first question to start with. (Note that we'd need additionally $m^2+n^2$ to be a square, too, and both conditions are by no means necessary, but they'd be sufficient.) While there are many squares of the form $m^2\pm mn+n^2,$ a brute force search doesn't yield any non-trivial solutions for $m^4+m^2n^2+n^4=k^2$ with $m,n<5\cdot 10^4$. Are there any?

Answer 1

This reduces to the quartic curve $y^2=x^4+x^2+1$ which has the (non-singular) rational point $(0,1)$, so is birational to an elliptic curve $E$ over $\mathbf{Q}$. Using the command in PARI/GP

E = ellfromeqn(y^2-x^4-x^2-1); E = ellinit(E); ellidentify(E)

one finds $E = 48a1$. Using the commands

elltors(E) and ellrank(E)

one sees that the Mordell-Weil group is $\mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z}$ so there is no non-trivial solution to the original equation.

Note that this is a bit circular since determining the rank of $E$ uses a $2$-descent and reduces to determine $E(\mathbf{Q})/2E(\mathbf{Q})$, thus solving a quartic equation as above.

Answer 2

Another proof that there is only the trivial solution is given in Mordell's book Diophantine Equations on page 19, equation (7'), by infinite descent. The crucial trick is to rewrite the equation as $(2k + 2m^2 + n^2)(2k - 2m^2 - n^2) = 3n^4$ and noting that the two factors are relatively prime (provided that $n$ is odd). Thus either \begin{align*} 2k + 2m^2 + n^2 &= u^4\\ 2k - 2m^2 - n^2 &= 3v^4, \end{align*} or \begin{align*} 2k + 2m^2 + n^2 &= 3u^4\\ 2k - 2m^2 - n^2 &= v^4, \end{align*} Modulo $16$ one gets a contradiction in the first case, and the second one quickly produces a smaller non-trivial solution.