Quintic diophantine equation

Pick any three numbers, say $1,2,3$. Compute $1^4+2^4+3^4=1+16+81=98$. Multiply through by $98^4$, and voila! $$98^4+196^4+294^4=98^5$$ If you insist on relatively prime solutions, you may have to work a little harder....

We can use the identity,

$$(2p-2q)^4+(2p+2q)^4+(4q)^4 = 2^5(p^2+3q^2)^2\tag1$$

One can then solve,

$$p^2+3q^2 = (a^2+3b^2)^k$$

for any $k$. For $k=5$, it is,

$$p =a^5 - 30 a^3 b^2 + 45 a b^4$$ $$q=b (5 a^4 - 30 a^2 b^2 + 9 b^4)$$

though $(1)$ has the common factor $2$.

Relatively prime may be difficult:

=======================

d       a       b       c
0       0       0       0
1       0       0       1
2       0       2       2
3       3       3       3
16       0       0      32
17       0      17      34
18      18      18      36
32       0      64      64
33      22      44      77
33      33      66      66
48      96      96      96
66     110     110     176

=======================