Problem 5. Find at least one solution to the equation

$$
a^{3}+b^{4}=c^{5}
$$

in natural numbers.
Solution. What can be taken as a working hypothesis, following which it will be possible to outline ways to find a solution? Let's make a table:

| $n$ | $n^{3}$ | $n^{4}$ | $n^{5}$ |
| ---: | ---: | ---: | ---: |
|  |  |  |  |
|  | 1 | 1 | 1 |
| 2 | 8 | 16 | 32 |
| 4 | 27 | 81 | 243 |
| 5 | 64 | 256 | 1024 |
| 6 | 216 | 625 | 3125 |
| 7 | 343 | 1296 | 7776 |
| 8 | 512 | 2401 | 16807 |
| 9 | 729 | 6596 | 32768 |
| 10 | 1000 | 10100 | 59049 |
| 11 | 1331 | 14641 | 100000 |
| 12 | 1728 | 20736 | 161051 |
| 13 | 2197 | 28561 | 248832 |
| 14 | 2744 | 38416 | 371293 |
|  | 3375 | 50625 | 537824 |

From the table, it is clear, firstly, that the equation

$$
n^{3}+n^{4}=n^{5}
$$

has no natural solutions.

Indeed, from equation (2) it follows that $n^{3}(1+n- \left.-n^{2}\right)=0$, and this equation has no natural solutions.

Secondly, it is obvious that $a>c$ or $b>c$.

Thirdly, the numbers $n^{5}$ end with the same digit as the number $n$. The number $n^{4}$ ends with $1,5,0$ or $6$.

Fourthly, if $n$ is an odd number, then $n^{3}, n^{4}$ and $n^{5}$ are odd numbers. If $n$ is even, then $n^{3}, n^{4}, n^{5}$ are even numbers.

Suppose that $a=2 k, b=2 p, c=2 m$. Then we get the following sequence of equations:

$(2 k)^{3}+(2 p)^{4}=(2 m)^{5} ; 8 k^{3}+16 p^{4}=32 m^{5} ; k^{3}+2 p^{4}=4 m^{5}$.

From this it follows that $k=2 l$ and

$$
8 l^{3}+2 p^{4}=4 m^{5} ; 4 l^{3}+p^{4}=2 m^{5} \text { and so on. }
$$

Now it becomes clear that $a=2^{x}, b=2^{y}$, $c=2^{z}(x, y, z$ - natural numbers). From equation (1) it follows

$$
\begin{gathered}
2^{3 x}+2^{4 y}=2^{5 z} \text { or } \\
2^{5 z}\left(2^{3 x-5 z}+2^{4 y-5 z}=2^{5 z} .\right.
\end{gathered}
$$

From this $2^{3 x-5 z}+2^{4 y-5 z}=1$.

Since $x, y, z$ are natural numbers, the last equation is equivalent to the following system of equations:

$$
\left\{\begin{array}{l}
3 x-5 z=-1 \\
4 y-5 z=-1
\end{array}\right.
$$

From this $z=0,2(3 x+1)=0,2(4 y+1)$.

Now it is clear that $x=8, y=6, z=5$.

Thus, $a=2^{8}, b=2^{6}, c=2^{5}$.