I deal with the following equation system: $$k \left(k^4+5 k^2 m+5 m^2\right)=1 \\ k^2+4 m=l^2$$
Mathematica and Maple show that the solution of this system for the ${k,m}$ pair can not be represented in radicals, while the solution for $m,l$ can be.
The question I have is the following may I relay on Mathematica or Maple or this is irrelevant?
And if the Mathematica and Maple are correct may I assume that this system has no rational solutions based on the fact that the solutions can not be represented in radicals?
$$ k(k^4 + 5k^2m +5m^2) = 1 \\ k^2 + 4m = \ell^2$$
If $k$ and $m$ are integers, then $k = k^4 + 5k^2m + 5m^2 = \pm 1$.
If $k=1$, we get $1 + 5m + 5m^2 = 1 \iff 5m(m+1)=0 \iff m \in \{0, -1\}.$
Then $k^2 + 4m = \ell^2$ becomes \begin{cases} \ell^2 = 1 & \text{If $m = 0$} \\ \ell^2 = -3 & \text{If $m = -1$} \\ \end{cases}
If $k=-1$, we get $5m^2 + 5m + 2 = 0,$ which has no integer solutions.
So we get $(k,\ell,m) \in \{(1,1,0), (1, -1, 0) \}$
Rational Solutions?
If you let $\ell = x+y$ and $k=x-y$, then $k^2 + 4m = \ell^2$ becomes $m = xy$ and $k(k^4 + 5k^2m +5m^2) = 1$ becomes $x^5-y^5 = 1$, which has no nontrivial rational solutions.