On a^4 + b^3 = c^2

There are infinitely many primitive solutions to the Diophantine
equation
                   a^4 + b^3 = c^2 

To prove this, factor the equation  b^3 = c^2 - a^4  as

            b^3  =  (c + a^2)(c - a^2)

If a and c have opposite parity the two factors on the right are
coprime, so we have integers m,n such that

           m^3  =  c + a^2          n^3  =  c - a^2

It follows that we have a solution if we can find integers a,c
of the form
               c = (m^3 + n^3)/2       a^2 = (m^3 - n^3)/2

These values of a^2 and c are clearly integers (since m and n are both
odd), so the only constraint on m,n is that they make "a" an integer.
There are several ways of constructing an infinite family with this
property.  For example, if we restrict ourselves to "twin" values
of m and n, meaning that m=k+1 and n=k-1, the above equation for a^2
becomes
                 a^2  -  3k^2  =  1

which is a "Pell equation".  By the usual methods we have the solutions 
given by k = 0, 1, 4, 15, 56, 209, 780, ... and in general by the
recurrence
                k_j  =  4 k_{j-1}  -  k_{j-2}

Obviously these values of k alternate between odd and even.  The
odd values give even values of m and n, so those correspond to non-
primitive solutions, such as 26^4 + 224^3 = 3420^2, whereas the even
values of k give odd values of m and n, and the corresponding
solutions of the original equation are primitive.  For example,
with k=4 we have m=5, n=3, which gives the solution a=7,b=15,c=76.
The next primitive solution in this family is based on the Pell
solution k=56, which gives m=57,n=55 and so a=97,b=3135,c=175784.
The next such solution would be based on the even Pell solution
k=780, and so on.  We can create other infinite families in a
similar way by defining a linear relation between m and n.