Quintic diophantine equation How can I find non trivial primitive integer solutions, to the Diophantine equation $$a^4+b^4+c^4=d^5$$
Can anyone find me solutions to this equation? 
Or if possible a parametric equation that generates solutions?
I would appreciate any help
Ive also simplified it to finding coprime integer solutions greater then 1 to the equation,$$xyz(x^2+y^2+z^2)=1250w^5$$
I don't know if that helps at all.
 A: 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 $2$.
A: 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

=======================
A: $k=1000;for(a=1,k,for(b=a,k,for(c=b,k,if(ispower(a^4+b^4+c^4,5,&n),print([a,b,c,n])))))
[3, 3, 3, 3]
[14, 252, 266, 98]
[18, 18, 36, 18]
[22, 44, 77, 33]
[33, 66, 66, 33]
[83, 83, 249, 83]
[96, 96, 96, 48]
[98, 196, 294, 98]
[110, 110, 176, 66]
[124, 174, 298, 98]
[163, 489, 489, 163]
[226, 226, 339, 113]
[356, 534, 534, 178]
[729, 729, 729, 243]$
A: 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....
A: 
a parametric equation that generates solutions?

The expected number of integer solutions without common factor is finite, so no.
