On solutions to special eighth/fourth powers $a^8+b^4 = c^8+d^4$? By the Lander-Parkin conjecture, it is assumed there are no non-trivial integer solutions to either,
$$a^8+b^8+c^8 = d^8+e^8+f^8\tag1$$
$$a^8+b^8 = c^8+d^8\tag2$$
However, if we relaxed $(1)$ a bit, then,
$$1118^8 + 1937^8 + 2502045^4 = 455^8 + 1846^8 + 3200691^4\tag{3a}$$
and after removing the common factor $13^8$ (pointed out by Adam Bailey),
$$86^8 + 149^8 + 14805^4 = 35^8 + 142^8 + 18939^4\tag{3b}$$
and infinitely many other primitive solutions like it. Likewise, if we relaxed $(2)$ to either,
$$a^8+b^4 = c^8+d^4\tag4$$
$$a^8+b^4 = c^4+d^4\tag5$$
then are there non-trivial solutions to $(4)$ and $(5)$?
P.S. In response to Bailey's answer, I forgot to add that solutions must be primitive. Otherwise, $(5)$ is easily solved.

Solutions of (5) can easily be obtained from solutions of the 4-2-2 equation.  Given any integers $A,B,C,D$ such that:
$$A^4+B^4=C^4+D^4$$
we have:
$$A^8+(AB)^4=(AC)^4+(AD)^4$$
Taking for example the smallest 4-2-2 solution:
$$59^4+158^4=133^4+134^4$$
we have:
$$59^8+9322^4=7847^4+7906^4$$
Unfortunately there doesn't appear to be any similar way to infer solutions of (4).
