Diophantine Equation With Varying Exponents

Question

I am considering the following Diophantine Equation - the approach I tried became the study of too many different cases - so many that I left it and tried to find an easier way. I wonder if anyone could shed any further light on the matter. The equation is interesting since it has 3 different exponents. $$ a^2 +b^3 = c^4 $$ $a$, $b$, $c$, are any integers. The approach I mentioned began by subtracting the $a^2$ on both sides and factoring the difference of two squares - but as I said, it quickly became very tedious.

Answer 1

Here's an infinite family of solutions,

$$\big(4q^2(p^2-2)\big)^2+(4dq^2)^3 = (2pq)^4$$

where $p,q$ solve the Pell equation $p^2-d^3q^2=1$. If you want the "reversed" form,

$$p^4 + (q^2-1)^3 = (q^3+q)^2$$

where $p^2-3q^2=1$. And an amazing one by Enrico Jabara,

$$(2^{88}7^{47}n^4)^3 + (2^{92}7^{49}n)^4 + (2^{32}7^{17}n^4)^5 + (2^8 7^4n^4)^7 + (n^4-2^{52}7^{28})^8 = (n^4+2^{52}7^{28})^8$$

which he probably found by expanding,

$$(an^4)^3 + (bn)^4 + (cn^4)^5 + (dn^4)^7 + (n^4-e)^8 = (n^4+e)^8$$

collecting powers of $n$, and solving for $a,b,c,d,e$.

Answer 2

I thought it possible to write a solution without using equations Pell.

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

If you make this change.

$$p=tz(2zk^2+t)$$

$$s=tzk^2(2zk^2-t)$$

The result of such decision.

$$a=sp^3$$

$$b=2tzk^2p^2$$

$$c=kp^2$$

Where the number $t,z,k$ - integers and set us. You may need after you get the numbers, divided by the common divisor.