<s>[INST] Show that a given equation has no solutions in integers Prove that there are no integers $x$ and $y$ such that 
$$x^5-x^4y-13x^3y^2+13x^2y^3+36xy^4-36y^5=77.$$
Firstly setting $x=y$ gives LHS=0. This suggests that $x-y$ is a factor of it, factoring gives 
\begin{align*}
(x-y)(x^4-13x^2y^2+36y^4) &=(x-y)(x^2-9y^2)(x^2-4y^2)\\ &=(x-3y)(x-2y)(x-y)(x+2y)(x+3y)=77.
\end{align*}
Im not sure how to proceed.
 [/INST] By my comment above, we have that $(x-y)=\pm 1, \pm 7, \pm 11$ or $\pm 77$. Similarly, $(x-2y)=\pm 1, \pm 7, \pm 11$ or $\pm 77$. 
Case I: Assume $x-y=1=x-2y$, then $y=0$ and $x=1$, but then $(x-3y)(x-2y)(x-y)(x+2y)(x+3y)=1\neq 77$.
Case II: Assume $x-y=-1$ and $x-2y=1$, then $y=-2$ and $x=-3$, but then $x+3y=-9$ which is not a divisor of $77$.
Case III: Assume $x-y=7$ and $x-2y=1$, then $y=6$ and $x=13$, but then $x+2y=25$ is not a divisor of $77$.
And so on. In principle you can proceed in this fashion and exclude all $64$ cases. You can remove many cases by making some more observations.
You can also work in a reverse way. Notice that $77=1\cdot 77$ or $ 7\cdot 11$. (or $-1\cdot -77$ or $-7\cdot -11$). Thus either one factor is $77$ and all others are $1$. Or one factor is $7$, one other factor is $11$ and all others are $1$ (up to signs). I admit that this seems like a brute force method, but it is doable and will work.
</s>