Integer solutions to $a^4+b^7=11^{11}$ 
Determine the solution set of $a^4 + b^7 = 11^{11}$ with $a,b \in \mathbb{Z}$.

Hints would be appreciated. I have tried working modulo $5$ and have deduced that $a$ or $b$ must be multiples of $5$. Not sure how helpful that is though.
 A: Try working under modulus $29 = 4\times 7 + 1$, we have


*

*$x \stackrel{def}{=} \text{mod}(a^4,29) \in \{ 0, 1, 7, 16, 20, 23, 24, 25 \}$

*$y \stackrel{def}{=} \text{mod}(b^7,29) \in \{ 0, 1, 12, 17, 28 \}$.

*$\text{mod}(11^{11},29) = 10$.


By brute force, one can verify there is no way to pick a $x$ from first list
and a $y$ from second list to satisfy $x + y \equiv 10 \pmod {29}$.
As a result, the equation
$$a^4 + b^7 = 11^{11}$$
has no integer solutions.
Update
About the question "Why $29$?", the basic idea goes like this.


*

*If one want to show an equation of the form "$A + B = C$, $C$ given" doesn't have a solution over $\mathbb{Z}$, one tactic is to show the same equation doesn't have a solution over $\mathbb{Z}/p\mathbb{Z}$ for some integer $p$. We usually prefer $p$ to be a prime
(or a power of a prime) because we know more about their properties.

*In order for this to have a chance to work, the possible overcome of $\text{mod}(A,p)$ and $\text{mod}(B,p)$ should be as small as possible. 

*When $A$ has the form of a power $a^n$, one way to limit the possible outcome of
$\text{mod}(A,p)$ is to choose $p$ such that $n$ is a factor of $\varphi(p)$, the number of integers smaller than and co-prime to $p$. If we further want $p$ to be a prime, this reduces the requirement to $n$ divides $p - 1$. Same thing happens if $B$ happens to 
has the form of a power.
In our case, both $A$ and $B$ have the form of a power. This means we want to look for
a prime $p$ of the form $\;\;k\,\text{lcm}(4,7) + 1\;\;$ for some positive integer $k$. 
The first candidate for $p$ is $29$ and we are lucky that it works.
