Solve in $\mathbb Z$ the equation: $x^5 +15xy + y^5=1$ Solve in $\mathbb Z$ the equation:  $x^5 +15xy + y^5=1$
I tried: $x(15y+x^4)+y^5=1$
But don't have much ideas on how to continue, thanks!
 A: Let's find all pairs $(x,y) \in  \mathbb{Z} \times \mathbb{Z} $ that satisfies 
$$\tag{*}\label{main-poly} x^5 +15xy + y^5=1$$
It's obvious that $(1,0)$ and $(0,1)$ are solutions of $\eqref{main-poly}$ , because if one of $x,y$ equals $0$, then the other one must be equal to $1$. 
It is also obvious, that if $x,y \neq 0$, then one of them must be positive, and the other one negative. 
Without loss of generality (because of symmetry), we can assume that $x>0$ and $y<0$. Also, if the pair $(x,y)$ is a solution, the pair $(y,x)$ will also be the solution.

For $ x = 2 $ we got:
$$ 30y + y^5 = -31 $$
which is true only for $y = -1$, because for every integer $y<-1$ the left side of equation will be always less than $-31$.
Therefore $(-1,2)$ and $(2,-1)$ safisfies $\eqref{main-poly}$. 

For $ x=3 $ there is no solution, because we get 
\begin{gather}
243 + 45y + y^5 = 1 & \Longleftrightarrow  & y(y^4+45) = -242 = -2 \cdot 121 = -2 \cdot (45 + 76)
\end{gather}
and there is no such integer $y$ satisfying this.

Let's see what happens for $ x \ge 4$.
Firstly, let's assume that $ \quad |y| \ge x \quad  \Longleftrightarrow \quad  y \le -x . \quad  $ Then we have no solutions, because
$$ x^5 + 15xy + y^5 \ \le \ x^5 + 15xy - x^5 \ = \ 15xy \ <  \ 0 \ < \ 1. $$

In the other case, that is for $ \ \ \  |y| < x \  \Longleftrightarrow \ y > -x   \Longleftrightarrow \ x > -y , \ $ we have:

$$ x^5 + 15xy + y \cdot y^4 > x^5 + 15xy + yx^4 = x^4 (x+y) + 15xy $$
because $ \ y^4 < x^4 \ $ and $ \  y<0, \ $ thus $ \ y^5 > yx^4. \ $

Next, because $ \ x \ge 4 \ $ and $ \ |y| \le x-1, \ $ we have
$$  x^5 + 15xy + y^5 \ > \  x^4 (x+y) + 15xy \ \ge  \ x^4 (x+(-(x-1))) + 15xy \ = \ x^4 + 15xy = \\  = \  x(x^2 \cdot x+15y) \ \ge \ x(4^2 x + 15y) \ = \ x(x + 15x + 15 y) \ > \ x(x + 15(-y) + 15 y) \ = \\ = x^2 \ \ge \ 16 \ > \ 1 $$
so there cannot be $ \ x^5 + 15xy + y^5 \ = \ 1$. 
So, the only possible solutions of $(*)$ are: $ (1,0)$, $ (0,1)$, $ (-1,2)$, $ (2,-1)$, 
