text: # 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!
  
  • In the future, you can write \mathbb{Z} to get $\mathbb{Z}$ and x^{5} to get $x^{5}$. – Ian Mateus Jan 1 '14 at 21:07
  • To state the obvious: $x=0,\ y=1$ is a solution, and $x=1,\ y=0$ is as well. – Daniel Robert-Nicoud Jan 1 '14 at 21:11
  • Or even just \Bbb Z and x^5. The braces are only needed for more than one digit/letter/whatever. – dfeuer Jan 1 '14 at 21:11
  • @dfeuer In LaTeX, \Bbb is obsolete and the syntax with braces is recommended. – egreg Jan 1 '14 at 21:14
  • Discuss if $x$ and $-y$ can differ by more than 1. As integer solution, $x$ must divide $y^5-1$,... – Lutz Lehmann Jan 1 '14 at 21:14
  
  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: $(1,0)$, $(0,1)$, $(-1,2)$, $(2,-1)$,
  
  • If you ask why $\ |y| \le x-1, \$ it's because $\ 0 < |y| < x \$ and $x,y$ are integers. – Kusavil Jan 2 '14 at 0:03