Let $a + b + c = 0$. Then $c = -a - b$.

Substitute into $a^4 + b^4 + c^4$:
$$a^4 + b^4 + (-a - b)^4 = 50$$
Expand $(-a - b)^4$:
$$(-a - b)^4 = (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$
So:
$$a^4 + b^4 + [a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4] = 50$$
$$2a^4 + 2b^4 + 4a^3b + 4ab^3 + 6a^2b^2 = 50$$
Group terms:
$$2(a^4 + b^4) + 4(a^3b + ab^3) + 6a^2b^2 = 50$$
Note that $a^3b + ab^3 = ab(a^2 + b^2)$.

Let $s = a + b$, $p = ab$.
Since $a + b + c = 0$, $c = -s$.

Now, $a^2 + b^2 = (a + b)^2 - 2ab = s^2 - 2p$.

Also, $a^4 + b^4 = (a^2)^2 + (b^2)^2 = [a^2 + b^2]^2 - 2a^2b^2 = (s^2 - 2p)^2 - 2p^2$.

$a^2b^2 = (ab)^2 = p^2$.

Substitute into the equation:
$$2[(s^2 - 2p)^2 - 2p^2] + 4p(s^2 - 2p) + 6p^2 = 50$$
Expand $(s^2 - 2p)^2$:
$$(s^2 - 2p)^2 = s^4 - 4s^2p + 4p^2$$
So:
$$2[s^4 - 4s^2p + 4p^2 - 2p^2] + 4p(s^2 - 2p) + 6p^2 = 50$$
$$2[s^4 - 4s^2p + 2p^2] + 4ps^2 - 8p^2 + 6p^2 = 50$$
$$2s^4 - 8s^2p + 4p^2 + 4ps^2 - 8p^2 + 6p^2 = 50$$
$$2s^4 - 8s^2p + 4ps^2 + (4p^2 - 8p^2 + 6p^2) = 50$$
$$2s^4 - 8s^2p + 4ps^2 + 2p^2 = 50$$
Recall $s = a + b$, but $a + b + c = 0$, so $s = -c$. But $a$, $b$, $c$ are symmetric, so $s$ can be any real number such that $a + b + c = 0$.

But $a + b + c = 0$ implies $s + c = 0$, so $c = -s$.

Now, $ab + bc + ca = ab + b(-s) + a(-s) = ab - bs - as = ab - s(a + b) = ab - s^2$.
But $a + b = s$, so $ab + bc + ca = ab - s^2$.

Let $x = ab + bc + ca = ab - s^2$.

But $ab = x + s^2$.

Let us try to find $x$.

Let us try specific values. Since $a + b + c = 0$, let $a = x$, $b = y$, $c = -x - y$.

Then $a^4 + b^4 + c^4 = x^4 + y^4 + (-x - y)^4 = 50$.

Expand $(-x - y)^4$:
$$(-x - y)^4 = (x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$$
So:
$$x^4 + y^4 + [x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4] = 50$$
$$2x^4 + 2y^4 + 4x^3y + 4xy^3 + 6x^2y^2 = 50$$
Let $x = t$, $y = -t$, $c = 0$.
Then $a + b + c = t + (-t) + 0 = 0$.

$a^4 + b^4 + c^4 = t^4 + (-t)^4 + 0^4 = 2t^4$.
Set $2t^4 = 50$, $t^4 = 25$, $t = \pm \sqrt{5}$.

So $a = \sqrt{5}$, $b = -\sqrt{5}$, $c = 0$.

Then $ab + bc + ca = (\sqrt{5})(-\sqrt{5}) + (-\sqrt{5})(0) + (0)(\sqrt{5}) = -5 + 0 + 0 = -5$.

Thus, $ab + bc + ca = -5$.