Positive integer solutions to $a^3 + b^3 = c^4$ Let $n$ and $m$ be positive integers. We know that
$n^3 + m^3 = n^3 + m^3$
Multiply both sides by $(n^3 + m^3)^3$; on the LHS you distribute, and on the RHS you use power addition rule;
$(n^4 + nm^3)^3 + (m^4 + mn^3)^3 = (n^3 + m^3)^4$
So we get an infinite array of solutions with
$a = n^4 + nm^3$
$b = m^4 + mn^3$
$c = n^3 + m^3$
Are there any solutions to the equation (in the title) that cannot be expressed in this form?
 A: There are many other solutions $(a,b,c)$.  First, start with arbitrary integers $u,v>0$ (you can add the condition $\gcd(u,v)=1$ so as to produce non-overlapping of infinite families of solutions).  Then, write
$$u^3+v^3=\prod_{r=1}^k\,p_r^{t_r}\,,$$
where $p_1,p_2,\ldots,p_k$ are pairwise distinct prime natural numbers and $t_1,t_2,\ldots,t_k\in\mathbb{Z}_{> 0}$.  Then, solve for $x_1,x_2,\ldots,x_r\in\mathbb{Z}_{\geq 0}$ from the congruences
$$3x_r\equiv -t_r\pmod{4}$$
for all $r=1,2,\ldots,k$.  (Well, this is quite easy and you should get $x_r\equiv t_r\pmod{4}$ for every $r=1,2,\ldots,k$.)  Then, take
$$a:=u\,\prod_{r=1}^k\,p_r^{x_r}\,,\,\,b:=v\,\prod_{r=1}^k\,p_r^{x_r}\,,\text{ and }c:=\prod_{r=1}^k\,p_r^{\frac{3x_r+t_r}{4}}\,.$$

For example, take $u:=3$ and $v:=5$.  Then,
$$u^3+v^3=27+125=152=2^3\cdot 19\,.$$
Hence, we can take
$$a:=3\cdot2^{4\alpha+3}\cdot19^{4\beta+1}\,,\,\,b:=5\cdot 2^{4\alpha+3}\cdot19^{4\beta+1}\,,\text{ and }c:=2^{3\alpha+3}\cdot 19^{3\beta+1}\,,$$
where $\alpha,\beta\in\mathbb{Z}_{\geq0}$.
Peter's answer can also be generated this way.  First, start with $u:=13$ and $v:=14$.  Then, $$u^3+v^3=3^4\cdot 61\,.$$  This leads to
$$a:=13\cdot 3^{4\alpha}\cdot 61^{4\beta+1}\,,\,\,b:=14\cdot3^{4\alpha}\cdot 61^{4\beta+1}\,,\text{ and }c:=3^{3\alpha+1}\cdot 61^{3\beta+1}\,,$$
where $\alpha,\beta\in\mathbb{Z}_{\geq 0}$.  Peter's answer corresponds to $\alpha=0$ and $\beta=0$.
Even the OP's example starts from $u:=m$ and $v:=n$. Then, take each $x_r$ to be $t_r$, so that $$\prod_{r=1}^k\,p_r^{x_r}=\prod_{r=1}^k\,p_r^{t_r}=m^3+n^3\,.$$
This gives
$$a:=m\left(m^3+n^3\right)\,,\,\,b:=n\left(m^3+n^3\right)\,,\text{ and }c=m^3+n^3\,.$$

Furthermore, if $(a,b,c)$ is a solution, then $(s^{4l}a,s^{4l}b,s^{3l}c)$ is a solution for any positive integers $s$ and $l$.  You can get all solutions $(a,b,c)\in\mathbb{Z}_{>0}^3$ to the equation $a^3+b^3=c^4$ in this way.
A: $$793^3+854^3=183^4$$ and $$183$$ is not the sum of two cubes.
