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Find all integral pairs $(x,y)$ satisfying $$ x^3=y^3+2y+1.$$

My approach:

I tried to factorize $x^3-y^3$ as $$(x-y)(x^2 + xy + y^2)=2y+1,$$ but I know this is completely helpless. Please help me in solving this problem.

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    $\begingroup$ $(1,0)$ is one of the solutions. I don't see another. $\endgroup$
    – prog_SAHIL
    Aug 14, 2018 at 17:20
  • $\begingroup$ Related: math.stackexchange.com/questions/2396705/… $\endgroup$
    – nonuser
    Aug 14, 2018 at 20:45
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    $\begingroup$ This is an interesting question, but the contest-math tag always leaves one wandering about the origin. I would feel a lot better about leaving this open, if you could still my fears and explain that this is not from a running contest. Can you, for example, give a link to the contest www-site? $\endgroup$ Aug 17, 2018 at 15:46
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    $\begingroup$ This question is given by my teacher. And he said he took it from AMTI previous year question collection. $\endgroup$
    – Identicon
    Aug 17, 2018 at 15:48
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    $\begingroup$ @JyrkiLahtonen this question was from Bulgarian MO 1999 $\endgroup$ Nov 26, 2020 at 8:43

7 Answers 7

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Hint: if $y>0$, then $y^3< y^3+2y+1< (y+1)^3$, so the RHS expression cannot be a perfect cube. A similar idea works if $y$ is a small enough negative number, but some negative numbers close to $0$ (or indeed $0$ itself) can provide a solution.

Try to find a lower bound, and then check the remaining possible values.

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  • $\begingroup$ Got the answer . Thanks for the hint. $\endgroup$
    – Identicon
    Aug 14, 2018 at 17:45
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    $\begingroup$ If $y<0$ then $|y^3+2y+1|=|y|^3+2|y|-1$ is strictly between $|y|^3$ and $(|y|+1)^3.$ $\endgroup$ Aug 17, 2018 at 18:01
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Or you can write $x=y+z$ for some $z\in Z$ (that is $z=x-y$). So we get a quadratic equation on $y$: $$ 3y^2z+y(3z^2-2)+z^3-1=0$$ which has a discriminant a perfect square: $$d^2 = (3z^2-2)^2-12z(z^3-1)=-3z^4-12z^2+12z+4$$

So we have $$3z^4+12z^2\leq 12z+4$$ and this can not be true for a lot of integers $z$ (in fact only for $0$ and $1$)...

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Solution

Assume that $$x=y+t,~~~t \in \mathbb{Z}.$$

Then $$(y+t)^3=y^3+2y+1,$$ namely $$3ty^2+(3t^2-2)y+t^3-1=0.\tag1$$

If $t=0$, then $y=\dfrac{1}{2}$, which is absurd. Therefore, $t \neq 0$, which implies that $(1)$ could be seen as a quadratic equation with respect to $y$.

Consider the discriminant for $(1)$. $$\Delta=(3t^2-2)^2-4\cdot 3t \cdot (t^3-1)=-3(t^2+2)^2+12t+16.$$

If $|t|\geq 2$, then $$\Delta=-3(t^2+2)^2+12t+16<-3(t+2)^2+12t+16=4-3t^2<0.$$ Thus, $(1)$ has no real root.

Therefore, the possible values of $t$ are $t=\pm 1.$ Now, we may verify that $t=1$ is the only one solution. Under this case, $$x=1,y=0.$$

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There are already a lot of solutions but I want to point out that your factorization can be completed to a proof:

Assume that y is nonnegative. From $$x^3=y^3+2y+1$$ follows that

$$x^3>y^3$$

and therefore $$x>y.$$ This implies $$3y\le 3y^2\le 1 (y^2+y^2+y^2)\lt (x-y)(x^2 + xy + y^2)=2y+1$$ and further $$y\lt 1 .$$

For negative $y$ a similar argument holds.

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we first consider only $x> y\ge0$

suppose $x=y+a$ then the equation is $(y+a)^3=y^3+2y+1$

$y^3+3ay^2+3a^2y+a=y^3+2y+1$

$3ay^2+(3a^2-2)y+(a-1)=0$

however, as $a\ge1$, $3a^2-2\ge0$,$a-1\ge0$

If $y>0$, then $3ay^2+(3a^2-2)y+(a-1)>0$, so $y=0$

then, $x^3=1$

$(x,y)=(1,0)$ are only integral solution

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  • $\begingroup$ But, $(-1,0)$ is NOT a solution. $\endgroup$ Aug 15, 2018 at 7:09
  • $\begingroup$ @SinTan1729 you are right, thanks $\endgroup$ Aug 16, 2018 at 2:45
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We have $$(y^3+2y+1)-(y-1)^3=3y^2-y+2=\frac{(6y-1)^2+23}{12}>0$$ and $$(y+2)^3-(y^3+2y+1)=6y^2+10y+7=\frac{(6y+5)^2+17}{6}>0\,.$$ That is, $$(y-1)^3<y^3+2y+1<(y+2)^3$$ for all $y\in\mathbb{Z}$. If $y^3+2y+1$ is a cube, then either $y^3+2y+1=y^3$ or $y^3+2y+1=(y+1)^3$, which gives $y=-\dfrac12$, $y=-\dfrac13$, or $y=0$. That is, $(x,y)=(1,0)$ is the only integer solution to $x^3=y^3+2y+1$.

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$$x^3=y^3+2y+1\tag{1}$$ We are after integer solutions of $(1)$. On setting $y=x-a$, where $a\in\mathbb{Z}$, we have from $(1)$ $$x^3=(x-a)^3+2(x-a)+1=x^3-3a x^2+(3a^2+2)x-a^3-2a+1\tag{2}$$ which rearranges to $$3a x^2-(3a^2+2)x+a^3+2a-1=0\tag{3}$$ Solve for $x$ \begin{align*} x&=\frac{(3a^2+2)\pm\sqrt{(3a^2+2)^2-12a(a^3+2a-1)}}{6a}\\ &=\frac{(3a^2+2)\pm\sqrt{-3a^4-12a^2+12a+4}}{6a}\tag{4} \end{align*} The discriminant is \begin{align*} -3a^4-12a^2+12a+4&=-3a^4-4(3a^2-3a-1)\\ &=-3a^4-4[3\left((a-\tfrac12)^2-\tfrac7{12}\right)]\\ &=-3a^4-12(a-\tfrac12)^2+7\\ &=7-(3a^4+3(2a-1)^2) \end{align*} Now check various integral $a$ for solutions: $a=0$ gives division by zero in $(4)$; $a=1$ gives $x_+=1$ or $x_-=\frac23$; $a=-1$ gives discriminant of $-23$; and for $|a|\ge2$ we get the discriminant $7-(3a^4+3(2a-1)^2)<0$ (since $3a^4+3(2a-1)^2$ is nonnegative).

Hence $x=1$, $y=0$ is the only solution.

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  • $\begingroup$ That is not a correct proof. You claim that $y^3+2y+1=(y-\alpha)^3$ for a fix alpha and all integer y. But there is no reason to claim this. Remove the restriction that x and y are integers an allow real values. Then you have infinite number of solutions but there exists no such alpha. $\endgroup$
    – miracle173
    Aug 22, 2018 at 5:42
  • $\begingroup$ @miracle173 I took the restriction that $x$ and $y$ were integers as read since we are after integer solutions. That's also why I didn't mention cube roots of unity, which I guess I should in that case. $\endgroup$ Aug 22, 2018 at 9:20
  • $\begingroup$ But restricting the solutions to integers still does not allow you to set $y^3+2y+1=(y-\alpha)^3$. Why should it? $\endgroup$
    – miracle173
    Aug 25, 2018 at 18:51
  • $\begingroup$ @miracle173 $y$ has to be an integer as we are after integer solutions. So $x=(y^3+2y+1)^{1/3}\in\mathbb{Z}$. Hence $y^3+2y+1$ is a cube, namely $x^3=(y-\alpha)^3$ on factorizing, i.e., $x=(y-\alpha)$. $\endgroup$ Aug 25, 2018 at 19:10
  • $\begingroup$ Of course you can define a function $\alpha(x,y)=x-y$ and write $y^3+2y+1=(y-\alpha(x,y))^3=y-3\alpha(x,y) y^2+3\alpha(x,y)^2y-\alpha(x,y)^3$ but on the right hand side there is no polynomial, because $\alpha$ is a not a constant, so you cannot compare the coefficients. $\endgroup$
    – miracle173
    Aug 25, 2018 at 19:44

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