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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!

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  • $\begingroup$ In the future, you can write \mathbb{Z} to get $\mathbb{Z}$ and x^{5} to get $x^{5}$. $\endgroup$
    – Ian Mateus
    Jan 1, 2014 at 21:07
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    $\begingroup$ To state the obvious: $x=0,\ y=1$ is a solution, and $x=1,\ y=0$ is as well. $\endgroup$ Jan 1, 2014 at 21:11
  • $\begingroup$ Or even just \Bbb Z and x^5. The braces are only needed for more than one digit/letter/whatever. $\endgroup$
    – dfeuer
    Jan 1, 2014 at 21:11
  • $\begingroup$ @dfeuer In LaTeX, \Bbb is obsolete and the syntax with braces is recommended. $\endgroup$
    – egreg
    Jan 1, 2014 at 21:14
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    $\begingroup$ Discuss if $x$ and $-y$ can differ by more than 1. As integer solution, $x$ must divide $y^5-1$,... $\endgroup$ Jan 1, 2014 at 21:14

1 Answer 1

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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)$,

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  • $\begingroup$ If you ask why $ \ |y| \le x-1, \ $ it's because $ \ 0 < |y| < x \ $ and $x,y$ are integers. $\endgroup$
    – Kusavil
    Jan 2, 2014 at 0:03

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