Find all integers $x,y$ which satisfy the equation $xy=20-3x+y$.
Problem
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Tags: number theory, Diophantine equation
geometrylover123
28.04.2021 18:23
Firstly, we have $\displaystyle xy-y=y( x-1) =20-3x$ Then we just have to solve this division equation $\displaystyle y=\frac{20-3x}{x-1} \in \mathbb{Z}$
SK1729
28.04.2021 18:30
xy=20-3x+y xy+3x-y-3=17 x(y+3)-1(y+3)=17 (y+3)(x-1)=17=can be 17*1 or 1*17 or (-1)*(-17) or (-17)*(-1) from this we get the soln as (y,x)=(14,2),(-2,18),(-20,0),(-4,-16).
SK1729
28.04.2021 18:32
NCEE wrote: Attachment which application/website is this?
TuZo
28.04.2021 18:33
geometrylover123 wrote: Firstly, we have $\displaystyle xy-y=y( x-1) =20-3x$ Then we just have to solve this division equation $\displaystyle y=\frac{20-3x}{x-1} \in \mathbb{Z}$ Continuation: $y=\frac{20-3x}{x-1}=\frac{3-3x+17}{x-1}=-1+\frac{17}{x-1}\in \mathbb{Z}\Rightarrow x-1\in \left\{ \pm 1,\pm 17 \right\}$
Lamboreghini
28.04.2021 18:47
SK1729 · 5 minutes ago (view)helo
NCEE wrote:
Attachment
which application/website is this?
#6hellloolo x 5 hellloolo Y 0 hellloolo
@SK1729 AoPS User Desmos.
jasperE3 · 24 minutes ago (view)helo
Find all integers $x,y$ which satisfy the equation $xy=20-3x+y$.
#1hellloolo x 5 hellloolo Y 0 hellloolo
@jasperE3 AoPS User This is a Diophantine equation, perfect for SFFT.
Begin by adding $3x-y$ to both sides of the equation to get $$xy+3x-y=20.$$We would like to factor this. Note that expanding $(x-1)(y+3)$ gives $xy+3x-y-3.$ Subtracting $3$ from both sides of the equation gives $$x+3x-y-3=20-3.$$Factoring gives $$(x-1)(y+3)=17.$$The only possible values for $x-1$ are $-17, -1, 1, 17.$ The corresponding $y+3$ values are $-1, -17, 17, 1.$ So we have
\begin{align*}
(x-1, y+3)&=(-17,-1),\\
(x-1, y+3)&=(-1, -17),\\
(x-1, y+3)&=(1,17),\\
(x-1, y+3)&=(17, 1).
\end{align*}
Simplifying gives the solutions
\begin{align*}
(x, y)&=\boxed{(-16, -4)},\\
(x, y)&=\boxed{(0, -20)},\\
(x, y)&=\boxed{(2, 14)},\\
(x, y)&=\boxed{(18, -2)}.
\end{align*}
SK1729
01.06.2021 10:38
Okay. Lamboreghini wrote:
SK1729 · 5 minutes ago (view)helo
NCEE wrote:
Attachment
which application/website is this?
#6hellloolo x 5 hellloolo Y 0 hellloolo
@SK1729 AoPS User Desmos.
jasperE3 · 24 minutes ago (view)helo
Find all integers $x,y$ which satisfy the equation $xy=20-3x+y$.
#1hellloolo x 5 hellloolo Y 0 hellloolo
@jasperE3 AoPS User This is a Diophantine equation, perfect for SFFT.
Begin by adding $3x-y$ to both sides of the equation to get $$xy+3x-y=20.$$We would like to factor this. Note that expanding $(x-1)(y+3)$ gives $xy+3x-y-3.$ Subtracting $3$ from both sides of the equation gives $$x+3x-y-3=20-3.$$Factoring gives $$(x-1)(y+3)=17.$$The only possible values for $x-1$ are $-17, -1, 1, 17.$ The corresponding $y+3$ values are $-1, -17, 17, 1.$ So we have
\begin{align*}
(x-1, y+3)&=(-17,-1),\\
(x-1, y+3)&=(-1, -17),\\
(x-1, y+3)&=(1,17),\\
(x-1, y+3)&=(17, 1).
\end{align*}
Simplifying gives the solutions
\begin{align*}
(x, y)&=\boxed{(-16, -4)},\\
(x, y)&=\boxed{(0, -20)},\\
(x, y)&=\boxed{(2, 14)},\\
(x, y)&=\boxed{(18, -2)}.
\end{align*}
OlympusHero
01.06.2021 19:59
Rewrite as $xy+3x-y=20$, or $(x-1)(y+3)=17$. The possibilities are $17 \cdot 1, 1 \cdot 17, -17 \cdot -1, -1 \cdot -17$. These correspond to $\boxed{(x,y)=(18,-2), (2,14), (-16,-4), (0,-20)}$.