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 $(4)$.
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).
$^1$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)
