Suppose a^2+b^3=c^4 for some prime numbers a, b, c.
Note that a=b=c=2 does not work, meaning for both sides of the equation to have the same parity, one of a, b, c must be 2, and the other two must be odd primes.
Next, note that we may subtract a^2 from each side of the equation to get
b^3 = c^4-a^2 = (c^2+a)(c^2-a)
=> b^2 = c^2+a and b = c^2-a
(as c^2+a > c^2-a and b is prime)
=> (c^2-a)^2 = c^2+a
We now consider three cases.
Case 1: a = 2
=> (c^2-2)^2 = c^2+2
=> c^4-4c^2+4 = c^2+2
=> c^4-5c^2+2 = 0
=> c^2 = (5+-sqrt(17))/2
=> c !in NN
but this contradicts the premise that c is a prime.
Case 2: b = 2
=> {(c^2-a = 2),(c^2+a = 4):}
=> (c^2-a)+(c^2+a) = 2+4
=> 2c^2 = 6
=> c^2 = 3
=> c != NN
again contradicting the premise that c is prime.
Case 3: c = 2
=> a^2+b^3=16
If a and b are both odd primes, then the least the left hand side can attain is when both a and b are the least odd prime, i.e. a=b=3, giving a^2+b^3 >= 3^2+3^3 > 16, a contradiction.
As each case leads to a contradiction, there are no three primes a, b, c satisfying a^2+b^3=c^4
