MHB Find all pairs (m,k) of integer solutions

[anemone]

Find all solutions to $(m^2+k)(m+k^2)=(m-k)^3$, where $m$ and $k$ are non-zero integers.

 

[Opalg]

Find all solutions to $(m^2+k)(m+k^2)=(m-k)^3$, where $m$ and $k$ are non-zero integers.

[sp]Multiply out the brackets: $$m^3 + m^2k^2 + mk + k^3 = m^3 -3m^2k + 3mk^2 - k^3$$, which (after dividing by the non-zero $k$) simplifies to $2k^2 + (m^2-3m)k + (m+3m^2) = 0.$ The discriminant of that quadratic in $k$ is $(m^2-3m)^2 - 8(m+3m^2)$, and that has to be a square, say $n^2$. So $$n^2 = (m^2-3m)^2 - 8(m+3m^2) = m^4 - 6m^3 - 15m^2 - 8m = m(m-8)(m+1)^2$$ and therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$. Thus $m = 9$ or $-1$. If $m=9$ then the quadratic for $k$ has solutions $k= -21$ and $k=-6$. If $m=-1$ then the only solution for $k$ is $k=-1$.

That gives three possible solutions, $(m,k) = (9,-21),\ (9,-6),\ (-1,-1)$.[/sp]

 

[I like Serena]

therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.

Why is that?

 

[Opalg]

therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.

Why is that?

If $x^2-16 = y^2$ then $16 = x^2-y^2 = (x+y)(x-y)$. The only factorisation of $16$ giving positive integer values for $x$ and $y$ is $x+y=8$, $x-y=2$, so that $x=5$ and $y=3$.

 

[anemone]

Thank you so much for participating, Opalg! And welcome back to the forum as well!:) But Opalg, I am sorry to tell you that you missed another solution set, which is $(m, k)=(8, -10)$.

My solution:

Spoiler

By expanding and simplifying the given equation, we get

$$m^3+m^2k^2+mk+k^3=m^3-3m^2k+3mk^2-k^3$$

$$2k^3+(m^2-3m)k^2+mk+3m^2k=0$$

Since $k \ne 0$, divide through the equation above by $k$, we have

$$2k^2+(m^2-3m)k+m+3m^2=0$$

and solve for $k$ using the quadratic formula yields

$$k=\frac{-(m^2-3m) \pm \sqrt{(m^2-3m)^2-4(2)(m+3m^2)}}{2(2)}$$

$$\;\;\;=\frac{3m-m^2 \pm \sqrt{(m+1)^2(m(m-8))}}{4}$$

$$\;\;\;=\frac{3m-m^2 \pm (m+1)\sqrt{m(m-8)}}{4}$$

Recall that the expression inside the radical must be greater than or equal to zero, thus, $$m(m-8)\ge 0$$ and this gives

$m\le 0$ and $m \ge 8$.

Also, bear in mind that we're told $m, k$ are both integer values, so, $m(m-8)$ has to be a squre.

But if we apply the AM-GM inequality to the terms $m$ and $m-8$, we find

$$\frac{m+m-8}{2} \ge \sqrt{m(m-8)}$$

$$m-4 \ge \sqrt{m(m-8)}$$

$m\ge 8$ implies $m-4 \ge 4$ and this further implies $\sqrt{m(m-8)} \le 4$ and now, we will solve for $m$ by considering $\sqrt{m(m-8)}=0$ or $1$ or $2$ or $3$or $4$ and we will take whichever solution that gives the integer values of $m$.

$$\sqrt{m(m-8)}= 4$$ gives irrational $m$ values.

$$\sqrt{m(m-8)}= 3$$ gives $m=-1, 9$.

$$\sqrt{m(m-8)}= 2$$ gives irrational $m$ values.

$$\sqrt{m(m-8)}= 1$$ gives irrational $m$ values.

$$\sqrt{m(m-8)}= 0$$ gives irrational $m=0, 8$ values but $m>0$, hence $m=8$.

Thus, by substituting each of the value of $m$ to the equation (*) above to find for its corresponding $k$ value, we end up with the following 4 solution sets, namely

$(m, k)=(-1, -1), (8, -10), (9, -6), (9, -21)$

 

[Opalg]

therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.

Why is that?

Oops, I should have taken I like Serena's query more seriously. I was thinking that the only way you could have $x^2-16 = y^2$ was in the situation $5^2 - 4^2 = 3^2$. But in the solution to this problem, the situation $4^2 - 4^2 = 0^2$ is also possible. That leads to the solution $m=8$ and $k=-10$.