Integer solution for $2^{3p}+2^{4q}=2^{5r}$ If $x,y,z$ are integer powers of $2$, prove that equation $$x^3+y^4=z^5$$
does not have solution, otherwise give a counterexample.
I tried to do this exercise, I firmly believe that the equation is not fulfilled, the detail is in how to show what is requested, can someone give me a hint please?
$$2^{3p}+2^{4q}=2^{5r}$$
We have:
$$2^{3p-4q}+1=2^{5r-4q}$$
$$\begin{align}\begin{cases} 3p-4q=0 \\5r-4q=1\end{cases}
&amp;\implies \begin{cases}3p=4q\\ 5r-4q=1\end{cases} \\
&amp;\implies 
\begin{cases} p=4m,q=3n \\ 5r-4q=1  \end{cases} \\
&amp;\implies \begin{cases}5r-12n=1 \\m =n.\end{cases}\end{align}$$
Applying Euclidean algorithm, we can find a counter-example:
$$p=28, q=21, r=17 $$ which is correct.
$$2^{84}+2^{84}=2^{85}$$
The only way to get a power of $2$ by adding two powers of $2$ is when the exponents are equal. Hence
$$3p=4q=5r-1.$$
As $3,4$ are relative primes, $p=4t,q=3t$ and
$$12t=5r-1.$$
The smallest solution is $t=2,r=5.$

$$256^3+64^4=32^5.$$