Expand both sides, cancel $m^3$ and divide by $n$ ($\neq 0$) to obtain $2n^2+(m^2-3m)n+3m^2+m=0$ $n_{1,2}={-m^2+3m\pm\sqrt{m^4-6m^3-15m^2-8m}\over 4}$ Factorize the discriminant as $m(m-8)(m+1)^2$. Therefore for some integer $k$ we must have $m(m-8)=k^2\iff (m-4)^2-k^2=16$. Thus $m-4\in\left\{\pm{1+16\over 2},\pm{2+8\over 2},\pm{4+4\over 2}\right\}\implies m\in\{-1,8,9\}$ (value $m=0$ is discarded). Plugging all of those into the expression for $n$, we find the solutions: $(m,n)\in\{(-1,-1),(8,-10),(9,-6),(9,-21)\}$

^Wait, why did you discard $m = -1$? I found that $(n + 1)(n^2 - 1) = -(n + 1)^3$ became $2n^3 + 4n^2 + 2n = 0$, yielding $n = -1$ and $n = 0$. So $\boxed{(-1,-1)}$ should be an additional solution...

Because I can't multiply In Farenhajt algebra, $-(-1)^2=1$. Thanks, professordad. Edited.

Reminds me of https://artofproblemsolving.com/community/c6h546190p3160606.

Note that $(m,0)$ is always a solution, so henceforth we assume $n\ne 0$. Expanding the equation, we get \[m^2n^2+mn+n^3=-3m^2n+3mn^2-n^3,\]or \[2n^2+n(m^2-3m)+(3m^2+m).\]The quadratic formula then gives \[n=\frac{3m-m^2\pm\sqrt{(m^2-3m)^2-8(3m^3+m)}}{4}=\frac{3m-m^2\pm(m+1)\sqrt{m(m-8)}}{4}.\]Thus, $m(m-8)$ is a perfect square, so $m\in\{-1,0,8,9\}$. This gives the final solution set of \[\boxed{(m,0),(-1,-1),(8,-10),(9,-6),(9,-21)}.\]

$n=0$ is clearly a solution so suppose $n\ne 0$; then, expanding and dividing by $n$ we get the quadratic $2n^2+(m^2-3m)n+(3m^2+m)=0$. But the discriminant is $m(m+1)^2(m-8)$ so $m^2-8m=x^2$, and thus $16+x^2=y^2$, so using diff. of squares we get $x^2=0, 9$. After a bit of bashing our final answer is $(m, n)=(m, 0), (-1, -1), (8, -10), (9, -6), (9, -21)$.

First, assume $a=b$. Then, we have $(a^2+a)^2=0\implies a=0\text{ or } -1$. This leads to the solutions $(0,0)$ and $(-1,-1)$. Now, assume $a=0$. We get $b^3=-b^3\implies b = 0$, which we already found. Finally, assume $b = 0$, which leads to $a^3=a^3$. Thus, all solutions of the form $(t,0)$ satisfy the equation. Now, assume both $a$ and $b$ are nonzero. Expanding the given equation gives us $$a^3+b^3+a^2b^2+ab=a^3-3a^2b+3ab^3-b^3$$$$\implies 2b^2+(a^2-3a)b+(a+3a^2)=0$$$$\implies b = \frac{3a-a^2\pm\sqrt{a^4-6a^3-15a^2-8a}}{4}.$$Since $b\in\mathbb{Z}$, we must have $\sqrt{a^4-6a^3-15a^2-8a}\in\mathbb{Z}$ as well. Notice that $a^4-6a^3-15a^2-8a=a(a-8)(a+1)^2$, so it is sufficient to find all $a$ such that $a(a-8)=k^2$ for some integer $k$. $$a^2-8a-k^2=0$$$$\implies a=\frac{8\pm\sqrt{64+4k^2}}{2}=4\pm\sqrt{16+k^2}.$$Now, we have to find all $k$ such that $16+k^2$ is a perfect square. Let $16+k^2=m^2$, so that $m^2-k^2=16$. Note that if $k\ge8$, then $$m^2-k^2\ge (k+1)^2-k^2=2k+1>17,$$so it is sufficient to check $k$ in the range $[0,7]$. We find that the only $k$ are $0,3$, which means $a\in\{-1, 0, 8, 9\}$. For each of these, we can manually check what $b$ can be, and we find the solutions $(-1,-1),(0,0),(8,-10),(9,-6),(9,-21)$. Remembering that the problem asks for nonzero solutions, the only possible pairs are $\boxed{(-1,-1),(8,-10),(9,-6),(9,-21)}$.

Expand to get \[m^3+m^2n^2+mn+n^3=m^3-3m^2n+3mn^2-n^3\]Rearranging, we get \[n(2n^2-n(3m-m^2)+(3m^2+m))=0\]The $n=0$ will always gives solutions, so we will assume $n\neq 0$ from now on. Otherwise, by the quadratic formula we must have the determinant of the inner quadratic is a square, \[\Delta^2 = (3m-m^2)^2-8(3m^2+m) = m(m+1)^2(m-8)\]Thus, $(m)(m-8)$ is a square. Note that $\gcd(m,m-8)\leq 8$. Case 1 is when both $m$ and $m-8$ are squares, this happens when $m=-1,9$. Case 2 is when either term is 0, $m=0,8$. Case 3 is when $m$ and $m-8$ are both of the form $2k^2$, which has no solutions. We may now answer extract. Our solution sets are (m,n) equals \[(m,0), (8,-10),(-1,-1),(9,-6),(9,-21)\]

OlympusHero wrote:

I don't get how you got this part: Since $9^2-8^2 = 17 > 16$ and $8^2-7^2 = 15 < 16$, we test until $8+4=12$.

If you let $m-4=x$, then $x^2-16$ as well as $x^2$ are both perfect squares, so the perfect squares must differ by $16$. The difference between two consecutive perfect squares eclipses $16$ at $9^2-8^2$ and we have $8^2-7^2=15<16$, so we test all the way until $8+4=12$. Now does that make sense?

OlympusHero wrote: If you let $m-4=x$, then $x^2-16$ as well as $x^2$ are both perfect squares, so the perfect squares must differ by $16$. The difference between two consecutive perfect squares eclipses $16$ at $9^2-8^2$ and we have $8^2-7^2=15<16$, so we test all the way until $8+4=12$. Now does that make sense? yea it makes sense now, thanks

The only solutions are $\boxed{(m,0)}$ for any integer $m$, and also the additional solutions $\boxed{(-1,-1), ((8,-10), (9,-6), (9,-21)}$ We can expand the equation and get \[ m^2n^2 + 3m^2 n - 3mn^2 + mn + 2n^3 = 0 \] If $n=0$, then we get $(m,0)$. Now assume $n\ne 0$. Then dividing both sides by $n$ gives \[m^2 n + 3m^2 - 3mn + m + 2n^2 =0\] This is the same as \[2n^2 + (m^2 - 3m)n + 3m^2 + m = 0\] The discriminant of this quadratic, which is \[(m^2 - 3m)^2 - 8(3m^2 + m) = m(m-8)(m+1)^2\]must be a perfect square. So $m(m-8)$ is a square. If positive integers satisfy $a^2 - b^2 = 16$, then we have $(a-b)(a+b) = 16$. Assume $a$ and $b$ are nonnegative. We have $(a,b) = (5,3), (4,0)$. Since $(m-4)^2 - m(m-8) = 16$, we have $m(m-8)\in 0, 9$, so $m\in \{-1,0,8,9\}$. This gives the described solutions.

The solution set is $(-1, -1), (8, -10), (9, 21), (9, -6)$. By expansion and doing no further simplification, $$(n+3)m^2 + (1-3n)m + 2n^2 = 0.$$Observe that the discriminant of this expression with respect to $n$ must be a perfect square, hence $$\Delta = (m^2-3m)^2-8(m+3m^2) = m(m-8)(m+1)^2$$is a perfect square. This yields $m=-1, 0, 8, 9$, which implies the desired solution set.

HamstPan38825 wrote: The solution set is $(-1, -1), (8, -10), (9, 21), (9, -6)$. By expansion and doing no further simplification, $$(n+3)m^2 + (1-3n)m + 2n^2 = 0.$$Observe that the discriminant of this expression with respect to $n$ must be a perfect square, hence $$\Delta = (m^2-3m)^2-8(m+3m^2) = m(m-8)(m+1)^2$$is a perfect square. This yields $m=-1, 0, 8, 9$, which implies the desired solution set. um how did you solve for $m$?

Except in the edge case where $(m+1)^2 = 0$, it's equivalent for $m(m-8)$ to be a perfect square. If that square is $s^2$, say, you can write it as $(m-4)^2 - s^2 = 16$, and then you can factor the resulting difference of squares.

Note, that if we expand, we have $a^3+a^2b^2+ab+b^3=a^3-3a^2b+3ab^2-b^3 \implies a^2b^2+ab+2b^3=3ab^2-3a^2b \implies a^2b+a+2b^2=3ab-3a^2$, hence, we have $2b^2+(a^2-3a)b+3a^2+a=0.$ If we use the quadratic, formula we get $\frac{3a-a^2\pm (a+1)\sqrt{a(a-8)}}{4}$, we need, $a(a-8)=(a-4)^2-16$, to be a perfect, square, note, that $(n+1)^2-n^2>16$, for all $n>7$(integers), so we just test, out the solutions, here, to get $\boxed{(-1,-1), (8,-10), (9,-6), (9,-21)}.$ @megarine, note, that it says for nonnegative integers, $m,n$, so $(m,0)$, is not a valid solution.

@above you mean nonzero I was going off this problem, so it's ok

Overcomplication ftw Expanding, we have \[m^2n^2+mn+n^3=-n^3+3mn^2-3m^2n \iff 2n^2+3m^2+m^2n-3mn+m=0 \text{ if } n \neq 0.\]Grouping the terms in terms of $m$, we find that \[(n+3)m^2+(-3n+1)m+2n^2=0,\]and by the quadratic formula, we get that \[m=\frac{(3n-1)\pm \sqrt{-8n^3-15n^2-6n+1}}{2(n+3)}=\frac{(3n-1)\pm \sqrt{-(8n-1)(n+1)^2}}{2(n+3)},\]meaning that $1-8n$ is a perfect square. Letting $n=\frac{1-k^2}{8}$, rewriting $m$ in terms of $k$ gives us that \[m=\frac{\frac{-5-3k^2}{8}\pm \sqrt{k^2\left(\frac{9-k^2}{8}\right)^2}}{2\left(\frac{25-k^2}{8}\right)}=\frac{(-5-3k^2)\pm k(9-k^2)}{2(5+k)(5-k)}.\]We now take this into cases. First, we must consider the case where $k=\pm5$. In this case, note that $n$ would then be $-3$, giving us that \[(m^2-3)(m+9)=m^3+9m^2-3m-27=(m+3)^3=m^3+9m^2+27m+27,\]which only has a solution at $m=\frac{9}{5}$, which is not an integer. Therefore this case has no solutions. Our second case is if we consider \[m=\frac{(-5-3k^2)+k(9-k^2)}{2(5+k)(5-k)},\]for $k\neq \pm 5$. Note that $k^3+3k^2-9k+5=(k+5)(k-1)^2$, meaning that $m$ is actually equal to \[\frac{k^2-2k+1}{2(k-5)},\]which means that $k-5 \mid k^2-2k+1$. Note that since $k^2-2k+1$ has a remainder of $16$ when divided by $k-5$, we actually have that $k-5 \mid 16$. Therefore, the possibilities for $k$ in this case are $-11$, $-3$, $1$, $3$, $4$, $6$, $7$, $9$, $13$, and $21$. Checking all of these, we find that the only valid $k$ are $\pm 3$, $7$, $9$, and $13$, giving us the solutions $(-1,-1)$, $(9,-6)$, $(8,-10)$, and $(9,-21)$. Our final case is if we consider \[m=\frac{(-5-3k^2)-k(9-k^2)}{2(5+k)(5-k)},\]for $k\neq \pm 5$. Note that since $k^3-3k^2-9k-5=(k-5)(k+1)^2$, we have that $m$ is actually equal to \[\frac{k^2+2k+1}{2(k+5)},\]which means that $k+5 \mid k^2+2k+1$, which has a remainder of $16$ when divided by $k+5$, meaning that $k+5\mid 16$. However, this is symmetric to the second case, meaning that all solutions that could be found here have already been found. Therefore, the only solutions are $(-1,-1)$, $(9,-6)$, $(8,-10)$, and $(9,-21)$, and we are done.

Expanding both sides gets us: \[m^3 + n^3 + m^2n^2 +mn = m^3 - 3m^2n + 3mn^2 - n^3\] \[2n^3 + (m^2 - 3m)(n^2) + (3m^2 + m)(n) = 0 \] Dividing by $n$, and then using the Quadratic Formula gets us: \[ \frac{-m^2 + 3m \pm \sqrt{m^4 - 6m^3 - 15m^2 - 8m}}{4}\] \[\frac{-m^2 + 3m \pm \sqrt{m(m - 8)(m + 1)^2}}{4}\] So, for the discriminant to be a perfect square, $m(m - 8)$ must be a perfect square. If $m + 1 = 0$, then this still holds true for $m(m - 8)$, as $-1 /cdot -9 = 9 = 3^2.$ This also holds true for $m = 0$. \[m(m - 8) = k^2\] \[(m - 4)^2 - k^2 = 16\] \[(m - 4 + k)(m - 4 - k) = 16\] This gives us the solution set $m = (-1, 0, 8, 9).$ Plugging these values into the quadratic, we get the solution set: $(m, n) = (-1, 1), (0, 0), (8, -10). (9, -6), (9, -21)$.

We claim the solutions are $\boxed{(m,n)\in\{(-1,-1),(8,-10),(9,-6),(9,-21)\}}$. Expanding gives, $$m^3 + n^3 + m^2n^2 + mn = m^3 - 3m^2n + 3mn^2 - n^3$$This simplifies to $$2n^3 + m^2n^2 + 3mn^2 - 3m^2n + mn = 0$$This rearranges to $$2n^2 + (m^2 - 3m)n + (3m^2 + m) = 0 \text{ for $n \neq 0$}$$Now quadratic formula gives $$n = \frac{-m^2 + 3m \pm \sqrt{m^4 - 6m^3 - 15m^2 - 8m}}{4} $$where the discriminant factors as $m(m-8)(m+1)^2$. Then for some $k$ we have $m(m-8) = k^2$. This then becomes, \begin{align*} m^2 - 8m &= k^2\\ (m-4)^2 - k^2 &= 16 \end{align*}This then gives the claimed solution set.

I claim all of the solutions to this equation are $\boxed{(-1, -1), (9, -6), (9, -21), (8, -10)}$. First expand both sides and remove $a^3$ both sides from the equation to obtain \[ab + a^2b^2 + b^3 = -3a^2b + 3ab^2 - b^3.\]Dividing by $b$ on both sides ($b \neq 0$) as well as rearranging, we obtain the quadratic in $b$ \[2b^2 + b(a^2 - 3a) + (3b^2 + b) = 0.\]Now we must have the discriminant $\Delta \in \mathbb{N}$, as $b$ must be an integer. Now upon factoring we find \[\Delta = a(a + 1)^2(a - 8) \implies a + 1 = 0, \text{ or } a^2 - 8a = (a - p)^2, \text{ for } p \in \mathbb{N}.\] Case 1: ($a + 1 = 0$). Then $a = -1 \implies 2b^2 + 4b + 2 = 0 \implies \boxed{b = -1}$. Case 2: $\left(a(a - 8) = (a - p)^2 \right).$ Then \[a = \frac{1}{2} \left(\frac{p^2}{p - 4} \right) \implies p = -12, -4, 2, 6, 8, 12, 20 \implies \boxed{a = -1, 9, 8}.\]Now $a = -1$ was covered in the first case, so that $a = 8, 9$ yields $\boxed{b = -10}$ and $\boxed{b = -6, -21}$, as claimed. $\blacksquare$

I was given no non-zero condition The solutions are $(m,n) = (m,0), (-1,-1), (8,-10), (9, -6), (9,-21)$. Note that $n=0$ produces a valid solution for any $m$ and $m=0$ implies $n=0$. Henceforth, assume $m,n \neq 0$. Move everything to the LHS and expand: \begin{align*} (m^2+n)(m+n^2) - (m-n)^3 &= [m^3+m^2n^2+mn+n^3] - [m^3-3m^2n+3mn^2-n^3] \\ &= 2n^3 + n^2(m^2-3m) + n(3m^2+n) = 0 \\ &\iff 2n^2+(m^2-3m)n+(3m^2+n) = 0. \end{align*} Viewing this equation as a quadratic in $n$, it suffices to find all values of $m$ such that \[\Delta_n = (m^2-3m)^2-4 \cdot 2 \cdot (3m^2+n) = m(m-8)(m+1)^2\] is a perfect square. Realize that $m = -1$ produces a solution, but otherwise, we must have \[m(m-8)=k^2 \iff (m-4)^2-k^2 = 16.\] Difference of squares and some factor yields $m=8$ or $m=9$; the desired solution sets follow.

Switch variables to $m$, $n$, and allow for zero solutions. We can expand and regroup to get the cubic in $n$ \[n\left(2n^2+(m^2-3m)n+(3m^2+m)\right) = 0.\] One solution to this is $n=0$, and the other two are of the form \[n = \frac{-(m^2-3m) \pm (m+1) \sqrt{m(m-8)}}{4}.\] The only perfect square values attainable by $m(m-8)$ are 16 and 25. Substituting back in, we get our solutions \[\boxed{(m,0), (-1,-1), (8,-10), (9,-21), (9,-6)}. \quad \blacksquare\]

Solved with v4913 (kinda) Notice that we can expand this to \begin{align*} m^3 + m^2 n^2 + m^2 + m n^2 &= m^3 - 3 m^2 n + 3 m n^2 - n^3 \\ m^2 n^2 + m^2 + mn^2 &= -3m^2n +3mn^2 - n^3 \\ 2n^2 + n(m^2-3m) + (m+3m^2) &= 0 \end{align*} Since the discriminant of this quadratic must be a perfect square, we have that $(m^2-3m)^2 - 8(m+3m^2)$ must be a perfect square. This simplifies to $m(m-8)(m+1)^2$. This means that either $m=0$, $m=8$, $m=-1$, or $m(m-8)$ is a perfect square. In order for $m(m-8)$ to be a perfect square, notice that $(m-4)^2$ is $16$ away and also a perfect square, so then in that case $m=9$. Thus, our answers are $\boxed{(-1, -1), (8, 10), (9, -6), (9, -21), (k, 0) \forall k \in \mathbb{Z}}$.