On solutions to special eighth/fourth powers $a^8+b^4 = c^8+d^4$?

Question

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 relax $(1)$ a bit, $$1118^8 + 1937^8 + 2502045^4 = 455^8 + 1846^8 + 3200691^4\tag{3a}$$ then remove the common factor $13^8$ (pointed out by Adam Bailey), $$86^8 + 149^8 + 14805^4 = 35^8 + 142^8 + 18939^4\tag{3b}$$

In fact, there are infinitely many other primitive solutions like it by solving,

$$(a+b)^8+(a−b)^8+(4ab)^4=(c+d)^8+(c−d)^8+(4cd)^4$$

Question: Likewise, if we relax $(2)$ 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 below, I forgot to add that solutions must be primitive. Otherwise, $(5)$ is easily solved.

Answer 1

Extending on On the eighth powers $A^8+B^8=C^8+D^8$: You provided a link to Wroblewski's database

Running a Python script$^1$ against that text file, I get:

#1: line 3 has 1 square: 292^4 + 193^4 = 257^4 + 16^8
#2: line 16 has 1 square: 4849^4 + 58^8 = 4303^4 + 4288^4
#3: line 54 has 1 square: 17332^4 + 23^8 = 17236^4 + 6673^4
#4: line 218 has 1 square: 213672^4 + 116309^4 = 203349^4 + 392^8
#5: line 253 has 1 square: 276598^4 + 223^8 = 253814^4 + 203329^4
#6: line 433 has 1 square: 747633^4 + 465236^4 = 682161^4 + 784^8
#7: line 441 has 1 square: 872^8 + 251873^4 = 676769^4 + 598772^4
#8: line 496 has 1 square: 923339^4 + 292^8 = 922933^4 + 190984^4
Done 1420 lines.

So 8 out of 1420 primitive 4-tuples are of kind $(5)$, and there are none of kind $(4)$.

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In case it is useful, here are formulae that generate infinite families of such 4-tuples. See also Finding formula that solves $w^4+x^4=y^4+z^4$ over the integers:

$$\begin{align} f_1(a) &= a^{7} + a^{5} - 2 a^{3} + 3 a^{2} + a \\ f_2(a) &= a^{13} - a^{12} - a^{11} - 5 a^{10} - 6 a^{9} + 12 a^{8} + 4 a^{7} - 7 a^{6} + 3 a^{5} + 3 a^{4} - 4 a^{3} - 2 a^{2} + a - 1 \\ f_5(a) &= a^{13} + 27 a^{12} - 214 a^{11} - 186 a^{10} - 2481 a^{9} + 861 a^{8} - 2804 a^{7} - 972 a^{6} - 2481 a^{5} - 27 a^{4} - 214 a^{3} + 294 a^{2} + a + 3 \end{align}$$

Then the 4-tuples are $(^h\!f(a,b), {}^h\!f(b,-a), {}^h\!f(a,-b),{}^h\!f(b,a))$ where $^h\!f$ denotes the homogenized version of eiter function.

Feed in $(a,b)=1$ and notice that the resulting 4-tuple might have a common factor of 2 (found empirically).

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$^1$For reference, here is the script so you can play around with it:

#!/usr/bin/env python3.8

from math import isqrt # Needs Python 3.8 or higher
from urllib.request import urlopen
import sys, re

URL = "https://www.math.uni.wroc.pl/~jwr/422/422-10m.txt"

# Pattern for 4 decimal numbers separated by white-space.
line_pat = re.compile (r'\s*(\d+)\s+(\d+)\s+(\d+)\s+(\d+)\s*')

count = 0

def check4 (vals, lineno):
    """ Vals is a 4-tuple of values that satisfy a^4 + b^4 = c^4 + d^4.
        Work out which of them are squares, and neaty print the result.
    """
    assert len(vals) == 4
    assert vals[0]**4 + vals[1]**4 == vals[2]**4 + vals[3]**4

    # Tuple of square-roots if perfect square, 0 otherwise.
    qs = tuple ( isqrt(v) if v == isqrt(v)**2 else 0 for v in vals )

    # How many squares do we have?
    n_qs = sum (bool(q) for q in qs)

    if n_qs != 0:
        global count
        count += 1
        print ("#%d: line %d has %d square" % (count, lineno, n_qs),
               end = ": " if n_qs == 1 else "s: ")
        for i in range(4):
            print ("%s^%d%s" % (qs[i] if qs[i] else vals[i],
                                8 if qs[i] else 4,
                                (" + ", " = ", " + ", "\n")[i]),
                   end = "")

lineno = 0

# with open ("*.txt") as data:
with urlopen (URL) as data:
    for bline in data:
        lineno += 1
        # Lines are decoded as "bytes", thus decode them as UTF-8.
        line = bline.decode().strip()

        match = line_pat.match (line)
        if match:
            values = tuple ( int(match.group(i)) for i in range(1,5) )
            check4 (values, lineno)
        else:
            print ("unrecognized line:", line, file=sys.stderr)
            sys.exit (1)

print ("Done %d lines." % lineno)

Answer 2

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).