MHB Solve for positive integer solutions

[anemone]

Solve for positive integers $x$ and $y$ where $x^2+x=y^3+y^2+y$.

 

[Wilmer]

Solve for positive integers $x$ and $y$ where $x^2+x=y^3+y^2+y$.

Spoiler

Trick question?

Pretty sure there's no solution to that.

2x = -1 + SQRT[1 + 4y(y^2 + y + 1)]

Tu est cruel, Anemone!

 

[anemone]

Spoiler

Trick question?

Pretty sure there's no solution to that.

2x = -1 + SQRT[1 + 4y(y^2 + y + 1)]

Tu est cruel, Anemone!

Spoiler

No, that is a genuine contest problem and solver is expected to convince us why the above equation has no solutions. (Malthe)

Au fait, Je suis tout mignon et gentil.:p

 

[kaliprasad]

Spoiler

No, that is a genuine contest problem and solver is expected to convince us why the above equation has no solutions. (Malthe)

Au fait, Je suis tout mignon et gentil.:p

needs closure

 

[anemone]

Solve for positive integers $x$ and $y$ where $x^2+x=y^3+y^2+y$.

My solution:

Spoiler

Note that $x^2+x=x(x+1)$ is the product of two consecutive positive integers whereas $y^3+y^2+y=y(y^2+y+1)$ can never represent the product of two consecutive positive integers. So we can conclude the problem has no solution.

 

[topsquark]

My solution:

Spoiler

Note that $x^2+x=x(x+1)$ is the product of two consecutive positive integers whereas $y^3+y^2+y=y(y^2+y+1)$ can never represent the product of two consecutive positive integers. So we can conclude the problem has no solution.

Spoiler

I'm not saying you are wrong, anemone, but your argument doesn't hold for something like the case 8 * 9 = 4 * 18. No one says that the x function and y function have to give the same pair of factors.

-Dan

 

[Opalg]

Solve for positive integers $x$ and $y$ where $x^2+x=y^3+y^2+y$.

[sp]Write the equation as $x^2 - y^2 + x - y = y^3.$ The left side factorises, giving $(x-y)(x+y + 1) = y^3.$ Suppose that $p$ is a prime common divisor of $x-y$ and $x+y+1$. Then $p$ also divides their product $y^3$ and therefore $p$ divides $y$. So $p$ divides $(x-y)+2y = x+y$. It follows that $p$ cannot divide $x+y+1$, contradicting the initial assumption about $p$.

That contradiction shows that no such $p$ can exist, and therefore $x-y$ and $x+y+1$ are coprime. But their product is a cube, and so they must each individually be cubes, say $x-y = u^3$ and $x+y+1 = v^3$, where $u,v$ are integers with $v>u>0.$ Also, $u^3v^3 = y^3$ so that $y=uv.$

Thus $x-uv = u^3$ and $x + uv + 1 = v^3$. Hence $(u^3 + uv) + uv + 1 = v^3$, from which $$2uv+1 = v^3 - u^3 = (v-u)(v^2 + uv + u^2).$$ But $v-u \geqslant 1$ and hence $v^2 + uv + u^2 \leqslant 2uv+1.$ Therefore $$0 < (v-u)^2 = v^2 - 2uv + u^2 = (v^2 + uv + u^2) - 3uv \leqslant (2uv+1) - 3uv = 1 - uv <0.$$ That is another contradiction, and the conclusion this time is that no positive integer solution $(x,y)$ to the original equation can exist.[/sp]

 

[anemone]

[sp]Write the equation as $x^2 - y^2 + x - y = y^3.$ The left side factorises, giving $(x-y)(x+y + 1) = y^3.$ Suppose that $p$ is a prime common divisor of $x-y$ and $x+y+1$. Then $p$ also divides their product $y^3$ and therefore $p$ divides $y$. So $p$ divides $(x-y)+2y = x+y$. It follows that $p$ cannot divide $x+y+1$, contradicting the initial assumption about $p$.

That contradiction shows that no such $p$ can exist, and therefore $x-y$ and $x+y+1$ are coprime. But their product is a cube, and so they must each individually be cubes, say $x-y = u^3$ and $x+y+1 = v^3$, where $u,v$ are integers with $v>u>0.$ Also, $u^3v^3 = y^3$ so that $y=uv.$

Thus $x-uv = u^3$ and $x + uv + 1 = v^3$. Hence $(u^3 + uv) + uv + 1 = v^3$, from which $$2uv+1 = v^3 - u^3 = (v-u)(v^2 + uv + u^2).$$ But $v-u \geqslant 1$ and hence $v^2 + uv + u^2 \leqslant 2uv+1.$ Therefore $$0 < (v-u)^2 = v^2 - 2uv + u^2 = (v^2 + uv + u^2) - 3uv \leqslant (2uv+1) - 3uv = 1 - uv <0.$$ That is another contradiction, and the conclusion this time is that no positive integer solution $(x,y)$ to the original equation can exist.[/sp]

Thanks Opalg for participating and thanks for the smart solution!(Cool)