x^3-3xy^2+y^3=n has a solution in integers x,y

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x^3-3xy^2+y^3=n has a solution in integers x,y

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Source: IMO 1982, Day 2, Problem 4

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orl

#1 h Nov 11, 2005, 9:32 PM • 2 Y

Y by Adventure10, Mango247

Prove that if $n$ is a positive integer such that the equation $x^3-3xy^2+y^3=n$ has a solution in integers $x,y$ , then it has at least three such solutions. Show that the equation has no solutions in integers for $n=2891$ .

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#2 h Nov 11, 2005, 9:56 PM • 2 Y

Y by Adventure10, Mango247

If $x, y$ is a solution then so is $y-x$ , $-x$ . Hence also $-y$ , $x-y$ .
If the first two are the same, then $y = -x$ , and $x = y-x = -2x$ , so $x = y = 0$ ,
which is impossible, since $n >0$ . Similarly, if any other pair are the same.

$2891 = 2 (mod.9)$ and there is no solution to $x^3 - 3xy^2 + y^3 = 2 (mod 9)$ .
The two cubes are each $-1$ , $0$ or $1$ , and the other term is $0$ , $3$ or $6$ ,
so the only solution is to have the cubes congruent to $1$ and the other term congruent to $0$ .
But the other term cannot be congruent to $0$ , unless one of $x$ , $y$ is a multiple of $3$ , in which case its cube is congruent to $0$ , not $1$ or $-1$ .

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feliz

#3 h Jan 5, 2009, 11:44 PM • 3 Y

Y by Hedy, Adventure10, Mango247

$(x + \omega y)^3 = x^3 - 3xy^2 + y^3 + \omega(3x^2y - 3xy^2)$

But $a^3 = (a\omega)^3 = (a\omega^2)^3$ , so, if there is a solution, we obtain other two.

For $n = 2891$ , we see the equation $mod 11$ . We must have $x^3 - 3xy^2 + y^3 \equiv 2891 \equiv 0$ . If $11 \nmid x, y$ there is a solution for the equation $t^3 - 3t + 1 \equiv 0$ (take $t = x.y^{-1}$ ). Testing cases, we see this is impossible. So $11 \mid x, y$ . This implies $11^3|x^3 - 3xy^2 + y^3 = 2891$ , contradiction.

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#4 h Aug 31, 2012, 10:07 AM • 1 Y

Y by Adventure10

Feliz's solution is false, since 11 doesn't divide 2891...

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#5 h Jun 9, 2017, 4:40 PM • 2 Y

Y by Adventure10, Mango247

What's the motivation behind the transformation $(x,y)\to (y-x,-x)$ ? I understand why one would want to go from $(x,y)\to (ax+by,cx+dy)$ , but is there a better way to find $a,b,c,d$ instead of plugging that in, equating coefficients, and solving a system of equations?

This post has been edited 1 time. Last edited by claserken, Jun 9, 2017, 4:43 PM

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#6 h Jun 9, 2017, 4:50 PM • 1 Y

Y by Adventure10

claserken wrote:

What's the motivation behind the transformation $(x,y)\to (y-x,-x)$ ? I understand why one would want to go from $(x,y)\to (ax+by,cx+dy)$ , but is there a better way to find $a,b,c,d$ instead of plugging that in, equating coefficients, and solving a system of equations?

Just write $y=x+k$ and see what happens.

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#7 h Sep 27, 2017, 8:03 PM • 2 Y

Y by Adventure10, Mango247

I would like to show another way to prove the second part of the problem and to do this I will prove more general result In fact there is no solution to $n=4k+2$ suppose to the contrary that there is a solution $(x,y)$ then discuss the following cases:
1) $x$ is even then clearly $y$ is even and Hence :
$(x,y)=(2a,2b)$ so $8(a^3-3ab^2+b^3)=4k+2$ which is wrong.
2) $x$ is odd and Hence:
a) $y$ is even then $n$ must be odd which is wrong.
b) $y$ is odd wthen $n$ must be odd which is wrong.
so when $n=4k+2$ no solution but $1982=4*495+2$ Hence no sloution and we are done

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#8 h Jun 7, 2018, 7:09 AM • 2 Y

Y by Amir Hossein, Adventure10

Detailed Solution as #2:
WLOG, Assume that $x=y+m$ ,
Then we have,
$(y+m)^3-3y^2(y+m)+y^3=n$ $m^3-3y^2m+(-y)^3=n$ So, we see that if $(x,y)$ is solution then $(-y,x-y)$ is also a solution.
Similarly, letting $y=x+k$ gives.
$k^3-3x^2k+(-x)^3=n$ So, $(-x,y-x)$ is also a solution.
Claim 1: All the solution pairs are distinct
assume to the contrary that $(x,y)=(-y,x-y)\implies y=-2y \implies x=y=0$ .
Contradiction since $n$ is a positive integer.
This proves our claim.So, we must have three different solution.

[Side note: In the case of cubes check $\pmod{7}$ or $\pmod{9}$ for relevant results]
Suppose for the sake of contradiction there exists a solution pair $(x,y)$ in integers for equation:
$x^3-3xy^2+y^3=2891$ We have, $x^3 \equiv 0,1,8 \pmod{9}$ ,
Now we work on the basis of: $2891 \equiv 2 \pmod{9}$ .
Case I: $x,y$ are multiple of $3$ .
$x^3-3xy^2+y^3 \equiv 0 \pmod{9}$ Contradiction.
Case II: $x$ or $y$ is multiple of three.
$\implies x^3-3xy^2+y^3 \equiv 0 \pmod{9}$ Contradiction.
Case III: $x,y$ are integers not multiple of 3.
$\implies x^3-3xy^2+y^3 \equiv 3,6 \pmod {9}$ Contradiction.
So, there are no solutions when $n=2891$ . And, we are done. $\blacksquare$

This post has been edited 3 times. Last edited by IMO2019, Jun 7, 2018, 7:27 AM

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ZETA_in_olympiad

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ZETA_in_olympiad

#9 h Apr 17, 2022, 11:58 AM

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We can use quadratic residues for solving this.

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#10 h Apr 11, 2023, 3:33 AM

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For the first part of the problem, notice that if $(x, y)$ is a solution, then $(y-x, -x)$ and $(-y, x-y)$ are also solutions, which means that all solutions come in distinct triplets of three, which is enough.

For the second part, notice that $2891 = 7^2 \cdot 59$ . Now, setting $a=yx^{-1} \pmod 7$ , we have $a^3 - 3a^2 + 1 \equiv 0 \pmod 7.$ If $a^3 \equiv 1 \pmod 7$ , then $a^2 \equiv 3 \pmod 7$ is not a quadratic residue. If $a^3 \equiv -1 \pmod 7$ , then $3a^2 \equiv 0 \pmod 7$ , contradiction as $7 \nmid a$ . This shows that there are no solutions.

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Nguyen

#13 h Sep 4, 2024, 11:53 AM

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So do you really want the motivation?

It can be factorized into $(x-y\cdot2\cos40^\circ)(x-y\cdot2\cos80^\circ)(x+y\cdot2\cos20^\circ)$ .
If $x-y\cdot2\cos40^\circ=0$ , then $x:y=\sin80^\circ:\sin40^\circ$ ,
and $x:y:(x-y)=\sin80^\circ:\sin40^\circ:\sin20^\circ$ .

Therefore, if $f(x,y)=0$ , then $f(x-y,x)=f(y,y-x)=0$ ,
and since the minimal polynomials of $x-y\cdot2\cos40^\circ$ , etc. are cubic,
they must differ by at most a constant factor.

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