Diophantine eq.

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Source: European Mathematical Cup 2017

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472 posts

#1 h Jan 3, 2018, 2:13 PM • 1 Y

Y by Adventure10

Solve in integers the equation :
$x^2y+y^2=x^3$

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UzbekMathematician

141 posts

UzbekMathematician

#2 h Jan 3, 2018, 2:32 PM • 3 Y

Y by Adventure10, Mango247, DEKT

$y^2-x^2y-x^3=0 \implies \Delta =x^4+4x^3=A^2 \implies x^2(x^2+4x)=A^2 \implies \boxed{(0;0)}$
$x^2+4x=B^2 \implies (x+2)^2-B^2=4 \implies x=0, x=-4 \implies \boxed{(-4;-8)}$

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#3 h Jan 3, 2018, 3:34 PM • 3 Y

Y by Mr.Techworm, Adventure10, Mango247

UzbekMathematician wrote:

$y^2-x^2y-x^3=0 \implies \Delta =x^4+4x^3=A^2 \implies x^2(x^2+4x)=A^2 \implies \boxed{(0;0)}$
$x^2+4x=B^2 \implies (x+2)^2-B^2=4 \implies x=0, x=-4 \implies \boxed{(-4;-8)}$

Good solution !

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#4 h Jan 3, 2018, 3:44 PM • 2 Y

Y by Adventure10, Mango247

UzbekMathematician wrote:

$y^2-x^2y-x^3=0 \implies \Delta =x^4+4x^3=A^2 \implies x^2(x^2+4x)=A^2 \implies \boxed{(0;0)}$
$x^2+4x=B^2 \implies (x+2)^2-B^2=4 \implies x=0, x=-4 \implies \boxed{(-4;-8)}$

That's how I solved it

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gethd

#5 h Jan 3, 2018, 3:54 PM • 1 Y

Y by Adventure10

UzbekMathematician wrote:

$y^2-x^2y-x^3=0 \implies$ $\ldots$

Small typo. It needs to be $y^2+x^2y-x^3=0$ .

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#6 h Jan 3, 2018, 6:00 PM • 1 Y

Y by Adventure10

nevermind

This post has been edited 1 time. Last edited by JANMATH111, Apr 15, 2020, 2:39 PM

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#7 h Jan 3, 2018, 6:15 PM • 1 Y

Y by Adventure10

JANMATH111 wrote:

Solved it during the contest:
x=0 <==> y=0
now x and y are not 0. Let gcd(x, y)=d
Then x=dk and y=dl and gcd(k, l)=1
then plugging in that stuff gives k^2ld+l^2=k^3d after division by d^2
Now we see that d|l^2 so l^2=nd.
plugging that in again and dividing by d gives k^2l+n=k^3
now we see that k^2|n|l^2
so k and l are not coprime after all... THat means k=l=1 but that is also not a solution so x=y=0.

$(-4,-8)$ is also a solution.

This post has been edited 1 time. Last edited by User335559, Jan 3, 2018, 6:15 PM

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#8 h Jan 4, 2018, 1:51 PM • 2 Y

Y by Centralorbit, Adventure10

This is my solution on the contest, similiar with @UzbekMathematician

We can rewrite the equation as
$y^2+x^2y-x^3=0$
Since the coefficient of $y^2$ on $LHS$ is $1$ which is different from $0$ , we can solve this equation as a quadratic equation with $y$ as variable. For $y$ to be an integer, it's neccessary for the discriminant to be a perfect square, so $x^4+4x^3=x^2x(x+4)=m^2$ where $m$ is an integer. This happens if and only if $x(x+4)$ is a perfect square, so lets write $x(x+4)=l^2$ , where $l$ is an integer. So
$x^2+4x-l^2=0$ . For the same reason, we solve this equation, as a quadratic equation, where $x$ is a variable. For $x$ to be an integer it's neccessary for the discrimant to be a perfect square. So we have
$16+4l^2=k^2$ , where $k$ is an integer.
So, $16=(k-2l)(k+2l)$
By checking all cases:
$16=16\cdot 1, 16=8\cdot 2, 16=4\cdot 4, 16=2\cdot 8, 16=1\cdot 16$
And the cases where the factors are negative, e.g $16=(-16)\cdot (-1)$ ( I checked all the cases one by one on the contest ), and from all the solutions I got that $l=0$ , so $x(x+4)=0$ , so $x=0$ or $x=-4$ . By substituting on the original equation we get that the only solutions are
$(0,0)$ and $(-4,-8)$ .

This post has been edited 5 times. Last edited by User335559, Jan 4, 2018, 2:26 PM

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#9 h Jan 4, 2018, 3:45 PM • 1 Y

Y by Adventure10

$y(x^2+y)=x^3$ .
Thus $x=ky$ , where $k\in{\mathbb{Q}}$

Simplifying gives $y=y^2(k^3-k^2)$ which gives $(0,0)$ as solution

Otherwise we have $y(k^3-k^2)=1$ so $\frac{1}{k^3-k^2} \in{\mathbb{Z}}$

Let $k=\frac{p}{q}$ where $(p,q)=1$

Then $\frac{q^3}{p^2-p^2q} \in{\mathbb{Z}}$

$(p^2(p-q))\mid{q^3}$ but $(p,q)=1$

Case 1: $p=q-1$ $\frac{(1+p)^3}{-p^2}$ integer so $p=1, q=2$

Case 2: $p=q+1$

Note implies $p,q$ both negative so solution covered in case $1$

Thus $y=2x$

giving $(-4,-8)$ as well

Thus we've exhaused our solution set $\blacksquare$

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#10 h Mar 29, 2019, 3:37 AM • 2 Y

Y by Adventure10, Mango247

If $x=0$ so $y=0$ . Now $x,y$ different of zero.
Divide the equation $x^2$ so $y+\frac{y^2}{x^2}=x$ . Thus $y=kx$ . So $xk+k^2=x$ . Then $x=-\frac{k^2}{k-1}$ with $k$ different of zero. So the unique solution is $k=2$ . Then $(x,y)=(-4,-8)$ and $(0,0)$ .

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#11 h Apr 6, 2020, 5:57 PM

Y by

Hello. Obvious solution is $x=y=0$ , so we will find another solutions. We can factorize this $x^2y+y^2-x^3=0$ , now we will divide both sides with $x^2$ and we will get:
$\frac{y^2}{x^2}+y-x=0$ . We will solve this equation using quadratic formula for $y$ and using that $a=\frac{1}{x^2}, b=1, c=-x$ , so:
$y=\frac{-1 \pm \sqrt{1+\frac{4}{x}}}{\frac{2}{x^2}}$ Since we only have one solution for $y$ , it means that discriminant has to be $0$ , so:
$\sqrt{1+\frac{4}{x}}=0$ and finally $x=-4$ and $y=-8$ . So, solutions are: $(x,y)=(0,0);(-4,-8)$

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MR.1

#12 h Dec 16, 2024, 5:06 PM • 1 Y

Y by MIC38

$x^2y+y^2=x^3$ (1) ( $x=0$ and $y=0$ is true and let's say that $x$ , $y$ isn't 0)
Since $x$ and $y$ are integers we know that $x^2$ | $y^2$ which means $x$ | $y$
Let's say that $y$ = $ax$ and put this in (1)
We get $ax^3$ + $a^2x^2$ = $x^3$ and since $x$ isn't 0 divide both sides by $x^2$
We get $ax+a^2=x$ now $a$ | $x$ and let's say $x$ = $ab$ put this $x$ and we get:
$a^2b+a^2=ab$ now divide both sides by $a$ and we get:
$ab+a=b$ and from here $a$ | $b$ and $b$ | $a$ which means that $a=b$ (2) or $a=-b$ (3)
From (2) we get $(x;y)=(0;0)$
From (3) we get $-a^2+a=-a$ which is same as $a^2=2a$ so $a=2$ (different from0)
and from this we get $(x,y)=(-4,-8)$
$ANSWER:(x,y)=(0,0),(-4,-8)$

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#13 h Apr 11, 2025, 5:41 PM

Y by

$y^2+x^2y-x^3=0 \implies \Delta =x^4+4x^3=A^2 \implies x^2(x^2+4x)=A^2 \implies \boxed{(0;0)}$
$x^2+4x=B^2 \implies (x+2)^2-B^2=4 \implies x=0, x=-4 \implies \boxed{(-4;-8)}$

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Ianis

#14 h Apr 11, 2025, 7:08 PM

Y by

Observe that $(x,y)=(0,0)$ is a solution and that $x=0\iff y=0$ . Hence assume $x,y\neq 0$ .
Rewrite the given equation as $y^2=x^2(x-y),$ which gives that $x^2\mid y^2$ and therefore that $x\mid y$ . Hence $y=kx$ for some $k\in \mathbb{Z}\setminus \{0\}$ . This gives $k^2=x(1-k),$ so $1-k\mid k^2$ . However, $\gcd (1-k,k^2)=1$ , so $k=0$ or $1-k=1$ or $1-k=-1$ . The first two can't happen because $k\in \mathbb{Z}\setminus \{0\}$ , so $1-k=-1$ and hence $k=2$ , which gives $x=-4$ and $y=-8$ , which is a solution to the original equation.
Hence the solutions are $(x,y)=(0,0)$ and $(x,y)=(-4,-8)$ .

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