Is there a limited amount of integer solutions to $a^{4}+b^{3}=c^{2}$?

Question

So far I have found $a=1 , b=2 , c=3$. Are there any other whole number solutions to this equation?

Answer 1

This is the generalised Fermat equation $$ x^p+y^q=z^r $$ for $(p,q,r)=(3,4,2)$, where we are in the spherical case $$ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{13}{12}>1. $$ There we have infinitely many integer solutions, see for example here.

Answer 2

Suppose $a^2$ is a triangle number (it is well-known that there are an infinite number of squares that are also triangular - see here). Then:

$$a^2 = \frac{x(x+1)}{2}$$

Putting $b=x+1$ and rearranging:

$$2a^2=b^2-b$$

Putting $c=a^2+b$:

$$c^2-a^4=(a^2+b)^2-a^4=a^4+2a^2b+b^2-a^4=(b^2-b)b+b^2=b^3$$

Hence $a^4+b^3=c^2$, the smallest non-trivial examples being $1^4+2^3=3^2$ and $6^4+9^3=45^2$.

Answer 3

Equation, $(a^4+b^3=c^2)$ has parameteric solution:

$a=(pqw^3)$

$b=(2q^2w^3)$

$c=(p^2q^2w^4+2q^2w^5)$

$p=4k-3$

$q=3k-2$

$w=2k^2-1$

For, $k=-1$

$(a,b,c)=(35,50,1275)$

$35^4+50^3=1275^2$

Answer 4

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

$a=t^2(u-1)v$

$b=t^3(v^2)$

$c=t^4uv^2(3u-2)$

Where, $t=2u^2-1$

$v=(2u-1)$

For u=2, we have:

$147^4+3087^3=(172872)^2$