How many triples (a, b, c) of integers exists such that a^4 + b^3 = c^2

Question

Give positive numbers a, b and c are such that
ab + bc + ca + abc = 4,
What is the value of
1/a+2 + 1/b+2 + 1/c+2 = ?

Answer 1

a^4 + b^3 = c^2 --> c^2-a^4 = b^3 --> we can put it as (c-a^2)*(c+a^2)=b*b^2 -->
c-a^2=b--> a^2=c-b-----> (1)
c+a^2=b^2 -----> (2)
Considering (1) in (2) we have c+c-b=b^2 --> b^2+b-2c=0,
to solve it as quadratic equation, b=(-1+sqrt(1+8c))/2, it has infinite number of solutions concerning b&c
lets see what can a get, obviously it has much less solutions:
b=1--> c=1--> a=0
b=2--> c=3--> a=1
b=3--> c=6--> a^2=3, no integer solution
...
b=9--> c=45--> a=6
...
b=50--> c=1275--> a=35
...
b=289--> c=41905--> a=204
..
b=1682--> c=1415403--> a=1189
I got these numbers with the help of excel within first 3000 values of b.
So based on above there are infinite number of solutions