Integer solution for $2^{3p}+2^{4q}=2^{5r}$

Question

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?

Answer 1

$$2^{3p}+2^{4q}=2^{5r}$$

We have:

$$2^{3p-4q}+1=2^{5r-4q}$$

$$\begin{align}\begin{cases} 3p-4q=0 \\5r-4q=1\end{cases} &\implies \begin{cases}3p=4q\\ 5r-4q=1\end{cases} \\ &\implies \begin{cases} p=4m,q=3n \\ 5r-4q=1 \end{cases} \\ &\implies \begin{cases}5r-12n=1 \\m =n.\end{cases}\end{align}$$

Applying Euclidean algorithm, we can find a counter-example:

$$p=28, q=21, r=17 $$ which is correct.

$$2^{84}+2^{84}=2^{85}$$

Answer 2

This is especial property of number,$ 2 $,for equations with sum of two terms on one side:

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

If $a=b$ then $2\cdot 2^a=2^{a+1}=2^c$

Least common multiplier of 2 and 3 is 6. So those multiples of 6 can be considered as powers of 2 which also satisfy this condition:

$2\times(6k)+1\equiv 0\bmod 5$

For example:

$5^2=25=2\times12+1$

So we can have:

$2^{12=3\cdot 4}+2^{12=4\cdot3 }=2^{25=5\cdot 5}$

Which gives $p=3, q=4, r=5$ .

Or $5\times 17=85, 85-1=84=3\times 28=4\times 21$

which gives $p=23, q=21$ and $r=17$

Set of numbers for $5r$ is:

$s=\{25, 55, 85, \cdot\cdot\cdot 25+30k\}$