That was also Germany TSTST Day 2, #1. It's basically the same as that USAMO problem, use the substitution $d=x-y$, assume WLOG $x \geq y$ and then solve the quadratic equation in $x$ by checking the determinant.

Thanks Titu:(

hajimbrak wrote: Determine all pairs $(x, y)$ of positive integers such that $\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1$ yea, too easy for shortlist! here's my solution = let $x\ge y$ than we have $7x^2-13xy+7y^2=(x-y+1)^3$ now let $x-y=a$ and hence we get $7a^2+x(x-a)=(a+1)^3\Longrightarrow x^2-ax-a^3+4a^2-3a-1=0$ now as $x,y$ are positive int. so discriminant of above quadratic in $x$ must be perfect square. hence $D=4a^3-15a^2+12a+4=(4a+1)(a-2)^2=m^2$ so $4a+1=k^2$. and thus $x=\frac{k^2-1\pm k(k^2-9)}{8}$ and $y=x-\frac{k^2-1}{4} = \frac{k^2-1\pm k(k^2-9)}{8} - \frac{k^2-1}{4}$ so we get family of solution for different values of $k$. so we are done

And then, what's the final answer, aditya 21 ?

aditya21 wrote: $x=\frac{k^2-1\pm k(k^2-9)}{8}$ and $y=x-\frac{k^2-1}{4} = \frac{k^2-1\pm k(k^2-9)}{8} - \frac{k^2-1}{4}$ If he didn't miscalculate (I didn't check, but I assume it is right :p), that's the answer, kunny.

hajimbrak wrote: Determine all pairs $(x, y)$ of positive integers such that $\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1$ You can discard the $\pm$ for most values of $k$ (i.e. $k \ge 3$), and for other values of $k$ you need to do some individual case checking. If you forget the case checking, you will be docked one point. (Edit: For $k = 3$, you need additional casework. )

Let $d = x-y$ where we let $x \ge y$. Then, we have $$7d^2 + xy = (d + 1)^3$$And thus $xy = d^3 - 4d^2 + 3d + 1$. We also have $x - y = d$ and so the quadratic formula would yield $$x = \frac{d \pm \sqrt{d^2 + 4(d^3 - 4d^2 + 3d + 1)}}{2} = \frac{d \pm |d-2|\sqrt{4d+1}}{2}$$This would mean that $4d + 1 = 4k^2 + 4k + 1$ for some integer $k$, and so $d = k^2 + k$. Substituting would give $$x = \frac{k^2 + k \pm |k^2 + k - 2|(2k+1)}{2}$$Remember that we're assuming $d \ge 0$ here. We're assuming $x > 0$ so therefore since $|k^2+k-2|(2k+1)$ is always non-negative, we see that $$x = \frac{k^2 + k + (2k+1)|k^2 +k - 2|}{2}$$and if $k^2 + k - 2 \ge 0$ (which is true for $k \ge 1$), then $x = k^3 + 2k^2 - k - 1$ and $y = k^3 + k^2 - 2k - 1$. Notice that we need $y$ to be positive so $k^3 + k^2 - 2k - 1 > 0$ must be satisfied. In order for that to happen, we need to toss out $k = 1$. Because $k^3 + k^2 - 2k - 1 = (k-2)(k^2 + 3k + 4) + 7$, we see that $y > 0$ for all $k \ge 2$. Lastly, we deal with when $k = 0$. This would yield that $x = 1$, and thus $y = 1$ (because we have that absolute value sign). Therefore, solutions are $$(1, 1) \text{ and } (x, y) = (k^3 + 2k^2 - k - 1, k^3 + k^2 - 2k - 1) \text{ for } k \ge 2$$

quadratic formula bashing? triggered

fclvbfm934 wrote: $$(1, 1) \text{ and } (x, y) = (k^3 + k^2 - 2k - 1, k^3 - 3k - 1) \text{ for } k \ge 2$$ this solution seem wrong with k=2 ?

I got $(1,1)$ and $(k^3+2k^2-k-1,k^3+k^2-2k-1)$ for $k\ge 2$ fclvbfm934 wrote: And thus $xy = d^3 - 4d^2 + 3d + 1$. We also have $x - y = d$ and so the quadratic formula would yield $$x = \frac{-d \pm \sqrt{d^2 + 4(d^3 - 4d^2 + 3d + 1)}}{2} = \frac{-d \pm |d-2|\sqrt{4d+1}}{2}$$ shouldn't it be $$x = \frac{d \pm \sqrt{d^2 + 4(d^3 - 4d^2 + 3d + 1)}}{2} = \frac{d \pm |d-2|\sqrt{4d+1}}{2}$$

WLOG $x \ge y$. The equation rearranges to $7(x-y)^2+xy=(x-y+1)^3$. Let $u=x-y$, $v=xy$. Then $7u^2+v=(u+1)^3$, so $v=u^3-4u^2+3u+1$. But $v=(y+u)y=y^2+uy$, so \begin{align*} y^2+uy=u^3-4u^2+3u+1 &\implies y^2+uy-(u^3-4u^2+3u+1)=0 \\ &\implies y=\frac{ -u\pm \sqrt{u^2+4(u^3-4u^2+3u+1)}}{2}. \end{align*}Therefore, $u^2+4(u^3-4u^2+3u+1)=k^2$ for some $k$. We can factor the left hand side to get \[ (u-2)^2 (4u+1)=k^2 \implies 4u+1=\ell^2\]for some $\ell$. But clearly $\ell$ is odd, so let $\ell=2m+1$, so that \[ 4u+1=(2m+1)^2=4m^2+4m+1\implies u=m^2+m. \]If $m=0$, then $u=0$, so $x=y$. Then, plugging back into the original equation, we get $x=y=1$. Otherwise, $m\ge 1$, so $u\ge 2$. Then, \[ k=(u-2)\sqrt{4u+1}= (m^2+m-2)\sqrt{4m^2+4m+1} = (m+2)(m-1)(2m+1),\]whence \[ y=\frac{-m(m+1) \pm (m+2)(m-1)(2m+1)}{2}.\]Clearly, the $\pm$ must be a $+$ in order for $y$ to be positive. Then, we simplify to get \[ y = m^3+m^2-2m-1, \ x=y+u=m^3+2m^2-m-1.\]But $y > 0$, so $m^3+m^2-2m-1 > 0$. Since this is an increasing function, and since $m=2$ satisfies the inequality, it is satisfied by all $m \ge 2$. Hence, our full solution set is \[ (x,y)=(1,1), \ (m^3+2m^2-m-1, m^3+m^2-2m-1) \ \forall m\ge 2\]and permutations, since the original equation is symmetric.

WLOG $x \ge y$. The equation is symmetric so in the end we can just swap $x$ and $y$. Now we have $7(x - y)^2 + xy = (x-y+1)^3$, so substitute $p = x - y$ and $q=xy$. This gives $$7p^2 + q = (p + 1)^3 = p^3 + 3p^2 + 3p + 1,$$so $q = p^3 - 4p^2 + 3p + 1$. Now as $x - y = p$ and $xy = q$ we have $(x + y)^2 = (x - y)^2 + 4q = p^2 + 4q$, and solving for $x$ and $y$ gives $$x = \frac{\sqrt{p^2 + 4q} + p}{2} \quad \text{and} \quad y = \frac{\sqrt{p^2 + 4q} - p}{2}.$$Now $p^2 + 4q = p^2 + 4p^3 - 16p^2 + 12p + 4= 4p^3 - 15p^2 + 12p + 4 = (p-2)^2(4p +1)$. Thus we seek for $\sqrt{4p + 1}$ to be an integer. It is well known that this is the case only when $p = k^2 + k$, where $k \in \mathbb{Z}^{+}$. This becomes $2k + 1$. Now $$x = \frac{(k^2 + k - 2)(2k + 1) - k^2 - k}{2} \quad \text{and} \quad y = \frac{(k^2 + k - 2)(2k + 1) - k^2 - k}{2}.$$Simplifying gives $$(x, y) = (1,1); \quad (k^3 + 2k^2 - k - 1, k^3 + k^2 - 2k - 1)$$and all permutations. $\square$

A different solution from the aboves. Using substitution and factoring. Quote: Determine all pairs $(x, y)$ of positive integers such that\[\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.\]Proposed by Titu Andreescu, USA If $x=y,$ we have $7(x-y)^2+xy=1,$ so $xy=1$ which gives $(1,1)$ as a solution. Note that if $(p,q)$ is a solution, than $(q,p)$ is a solution. WLOG, let $x > y.$ Then, making the substitution $x-y=b,$ $x+y=a$ or $x=(a+b)/2$ and $y=(a-b)/2$ we get (after cubing) $$7\left(\frac{a+b}{2}\right)^2-13\left(\frac{a+b}{2} \right)\left(\frac{a-b}{2} \right) + 7\left(\frac{a-b}{2}\right)^2=(b+1)^3$$This reduces to $$a^2=4b^3-15b^2+12b+4$$We see that $a$ and $b$ must have same parity. If $a,b$ are both even, let $a=2m, b=2n,$ and $m > n.$ Substituting in gives $4m^2=4(8n^3)-15(4n^2)+12(2n)+4,$ so $m^2=8n^3-15n^2+6n+1,$ which miraculously factors as $\boxed{m^2=(n-1)^2(8n+1)}.$ Now we just want to find $n$ such that $8n+1$ is a square. However, $1^2,3^2,5^2,7^2$ are all congruent to one mod 8, so for any odd square number at least 9, we can get such an $n.$ So, if we let $n=\frac{(2c+1)^2-1}{8}=\frac{c^2+c}{2}$ for $c \geq 2,$ (If $c=1$ then $n=1, m=0,$ but we must have $m \geq n$) we have $m= \left(\frac{c^2+c}{2}-1\right) \cdot (2c+1)$ So, $a=(c^2+c-2)(2c+1)$ and $b=c^2+c.$ We plug these back into $(x,y)=((a+b)/2,(a-b)/2)=\boxed{(c^3+2c^2-c-1,c^3+c^2-2c-1)},$ which along with $\boxed{(1,1)},$ comprise the complete solution set. If $a,b$ are both odd, let $a=2m+1,b=2n+1.$ Substituting and factoring gives $(2m+1)^2=(2n-1)^2(8n+5),$ but $8n+5$ is never a square, so we are done. We can also take mod 4 to see that the RHS would be 15 = 3 mod 4, but squares are 0 or 1 mod 4. Oh also take permutations, of the solutions, because like we said earlier $(p,q)$ is a solution means $(q,p)$ is a solution.

WLOG, $x\ge y$. We first solve the easy case of $x=y$; we get $\sqrt[3]{x^2}=1$, so $x=y=1$. Now let $x-y=d>0$. Write \[\sqrt[3]{7x^2-13xy+7y^2}=d+1.\]Rewrite \[7x^2-13xy+7y^2=7(x-y)^2+xy=7d^2+y^2+yd.\]Cube both sides to yield \[7d^2+y^2+yd=d^3+3d^2+3d+1\implies y^2+yd-(d^3-4d^2+3d+1)=0.\]By the quadratic formula, we have \[y=\frac{-d\pm \sqrt{d^2+4(d^3-4d^2+3d+1)}}{2}.\]Factor \[4d^3-15d^2+12d+4=(d-2)(4d^2-7d-2)=(d-2)^2(4d+1).\]Noting that $y$ is positive, we write \[y=\frac{-d+|d-2|\sqrt{4d+1}}{2}.\]Note $d=1$ does not work because $\sqrt{4d+1}=\sqrt{5}$ in that case, so we have $d\ge2$, and \[y=\frac{-d+(d-2)\sqrt{4d+1}}{2}.\]Moreover, $d$ cannot be odd; otherwise, $4d+1\equiv5\pmod{8}$ implies $y$ is not an integer. So, write $d=2k$, so \[y=\frac{-2k+(2k-2)\sqrt{8k+1}}{2}=-k+(k-1)\sqrt{8k+1}.\]Now, note $8k+1$ must be an odd square because $\sqrt{8k+1}$ is rational and an algebraic integer, so it is a rational integer. Suppose $8k+1=(2\ell+1)^2$, then $4(\ell^2+\ell)+1=8k+1$ implies $k=\frac{\ell^2+\ell}{2}$. Now, we have \[y=-\frac{\ell^2+\ell}{2}+\left(\frac{\ell^2+\ell-2}{2}\right)(2\ell+1).\]For $\ell>1$, this expression is positive, so we get the curve of solutions \[(x,y)=\left(\frac{\ell^2+\ell}{2}+\left(\frac{\ell^2+\ell-2}{2}\right)(2\ell+1),-\frac{\ell^2+\ell}{2}+\left(\frac{\ell^2+\ell-2}{2}\right)(2\ell+1)\right)\forall\ell\in\mathbb{Z}_{>1}\]along with the pair $(1,1)$. Since we initially assumed $x\ge y$, we also have the curve of solutions \[(x,y)=\left(-\frac{\ell^2+\ell}{2}+\left(\frac{\ell^2+\ell-2}{2}\right)(2\ell+1),\frac{\ell^2+\ell}{2}+\left(\frac{\ell^2+\ell-2}{2}\right)(2\ell+1)\right)\forall\ell\in\mathbb{Z}_{>1}.\]

Knowing who wrote the problem made this quite a lot easier Note that if $x=y$, then the only solution is $(1,1)$. Now, we assume $x>y$. Do the titu substitution and let $a=x-y$ and $b=x+y$, therefore $a,b\in \mathbb{Z}_{>0}$. Now, the given identity becomes \[\frac{27a^2+b^2}{4}=7(x-y)^2+xy = (x-y+1)^3=(a+1)^3\]Rearranging, this becomes \[b^2 = 4a^3-15a^2+12a+4 = (a-2)^2\cdot (4a+1)\]Therefore, $4a+1 = (2k+1)^2$ for some integer $k$, so all solutions are characterized by $a=k^2+k$ and $b=(k^2+k-2)\cdot (2k+1)=(k-1)(k+2)(2k+1)$. This is obviously symmetric in $x,y$, and we have found all possible solutions. Thus, the only positive solutions are \[(a,b) =\left(k^2+k,(k-1)\cdot (k+2)\cdot (2k+1)\right)\]And $x,y$ follow from this, the solutions to $x,y$ are $(1,1)$ and also for all $k>0$ \[x=\frac{k^2+k+(k-1)(k+2)(2k+1)}{2}, y=\frac{(k-1)(k+2)(2k+1) - (k^2+k)}{2}\]$\blacksquare$.

IMC 2019

All solutions include $(x,y)=(1,1),(\frac{k^2+k+(k-1)(k+2)(2k+1)}{2}, \frac{(k-1)(k+2)(2k+1) - (k^2+k)}{2})$ Without the loss of generality,assume that $x>y$ Substitute $a=x-y$ and $b=x+y$ Now, the given identity becomes $\frac{27a^2+b^2}{4}=7(x-y)^2+xy = (x-y+1)^3=(a+1)^3$ Rearranging, this becomes $b^2 = 4a^3-15a^2+12a+4 = (a-2)^2\cdot (4a+1)$ Or $(4a+1)=(2k+1)^2$ for some $k \in \mathbb(Z)$ whose solutions are characterized by $a=k^2+k$ and $b=(k^2+k-2)\cdot (2k+1)=(k-1)(k+2)(2k+1)$,and solving for $(x,y)$ we get the answer.

IMC 2019

If $x=y$, we have the unique solution $(x,y)=\boxed{(1,1)}$. Let $s=x-y$, WLOG $s>0$. The equation rewrites as: $$7s^2+xy=(s+1)^3\Leftrightarrow s^2+y(y+s)=(s+1)^3\Leftrightarrow y^2+sy-s^3+4s^2-3s-1=0.$$Then the discriminant in $y$: $$\Delta=4s^3-15s^2+12s+4=(4s+1)(s-2)^2$$must be a square, so $4s+1=r^2$ for some $r>1$. This means that $s\ne1$, so $s\ge2$. Since $r\equiv1\pmod2$, $s=t^2+t$ where $r=2t+1$ and $t>0$. By the quadratic formula: $$y=\frac{-s\pm\sqrt\Delta}2=\frac{-t^2-t\pm r(s-2)}2=\frac{-t^2-t\pm(2t+1)(t^2+t-2)}2\in\{t^3+t^2-2t-1,-t^3-2t^2+t+1\}.$$Of these, $-t^3-2t^2+t+1$ is negative for $t\ge1$ and $t^3+t^2-2t-1$ is negative for $t=1$, so $(x,y)=\boxed{(t^3+2t^2-t-1,t^3+t^2-2t-1)}$ for some $t\ge2$ (we can check that these solutions work). Generalizing again, $\boxed{(t^3+t^2-2t-1,t^3+2t^2-t-1)}$ is the other set of solutions.

The solutions are $(1,1)$ and permutations of $(k^3+2k^2-k-1,k^3+k^2-2k-1)$ for $k \ge 2$. Note that these work. If $x=y$, we get $(1,1)$ and no other solutions. Assume $x>y$ and let $a=x-y$. Let $b=x+y$. We have that $$\frac{27a^2+b^2}{4} = 7x^2-13xy+7y^2=(a+1)^3 \implies b^2=4a^3-15a^2+12a+4 = (4a+1)(a-2)^2.$$This means $4a+1$ is a perfect square, so $a=k^2+k$ for some $k$. This means $b=(2k+1)(k^2+k-2) = 2k^3+3k^2-3k-2$. This gives the solutions $(x,y) = (k^3+2k^2-k-1,k^3+k^2-2k-1)$ for $k \ge 2$ as desired so we are done.

Let $d=x-y$ and WLOG $d\ge 0$. We get $y(y+d)=d^3-4d^2+3d+1.$ Thus, the discriminant wrt $y$, which is $4d^3-15d^2+12d+4=(d-2)^2(4d+1)$ is a perfect square. Since $4d+1$ is a square and is odd, we have $d=\frac{(2n+1)^2-1}{4}=n^2+n.$ Then, applying quadratic formula gives $y\in \{-n^3-2n^2+n+1,n^3+n^2-2n-1\}.$ Since $y$ is positive, $n\ge 2$ and $y=n^3+n^2-2n-1$ and $x=n^3+2n^2-n-1.$ Permutations also work. Is this not literally 98% same as USAMO 2015/1

$(x, y) = (k^3+2k^2-k-1, k^3+k^2-2k-1)$ or $(k^3+k^2-2k+1, k^3+2k^2-k-1)$ for $k \geq 2$ and $k = -1$ by expansion and positive condition. Take it or leave it.

The only solutions are $\boxed{(1,1), (k^3 + 2k^2 - k - 1, k^3 + k^2 - 2k - 1)}$, and permutations, for any integer $k\ge 2$. These work. Now we prove they are the only solutions. WLOG $x\ge y$ and let $a = x-y$. We can cube both sides of the equation and get \[7a^2 + x(x-a) = (a+1)^3,\]which can be rearranged to \[x^2 - ax -a^3 + 4a^2 - 3a - 1 =0\]Notice that $a=2$ implies that $x=1$ and $y=-1$, which is impossible, so assume $a\ne 2$ from now on. The discriminant of the equation, which is $a^2 + 4(a^3 - 4a^2 + 3a + 1) = (a-2)^2 (4a + 1),$ must be a perfect square, so $4a+1$ is also a perfect square. Let $4a+1 = (2k+1)^2$ for some nonnegative integer $k$. We get $a = k^2 + k$. By the quadratic formula, \[x = \frac{a\pm (a-2)(2k+1)}{2}\]If $a=0$, then obviously $(x,y) = (1,1)$, so now assume $a$ and $k$ are both positive. Since $a = k^2 + k$, clearly $a=1$ is impossible, so $a>2$ and $k>1$. The positive direction must hold in the equation for $x$, because $a < (a-2)(2k+1)$ and $x>0$. Thus, we can find \begin{align*} x \\ = \frac{a + (a-2)(2k+1)}{2} \\ = \frac{k^2 + k + (k^2 + k - 2)(2k+1)}{2} \\ = \frac{2k^3 + 4k^2 - 2k - 2}{2} \\ = k^3 + 2k^2 - k - 1 \\ \end{align*}Since $x-y = k^2 + k$, we get $y = k^3 + k^2 - 2k - 1$. Thus, our solutions claimed in the beginning are the only ones.

Kunihiko_Chikaya wrote: And then, what's the final answer, aditya 21 ? there doesnt need to be a list of ordered pairs $(a_1, b_1) \cdots {(a_i, b_i)}.$ As long as you can express all solutions in $k$ format such that $k\in \mathbb{R}$ (and even that may not be necessary), you're done.

Note that the statement is clearly symmetric in $x$ and $y$, so suppose WLOG that $(x,y)$ is a solution with $x\ge y$. We let $d=x-y\ge 0$, so that the problem statement rewrites as \[\sqrt[3]{7d^2+xy}=d+1\]\[\iff xy=(d+1)^3-7d^2=d^3-4d^2+3d+1.\]It follows that \[(x-1)(y+1)=xy+d-1=d(d-2)^2,\]or equivalently \[a(a+t)=t^2(t+2)\quad (*),\]where we let $t=d-2$ and $a=y+1$. Note that $a>0$ and thus $t\ne 0$. It makes sense now to consider the gcd of $a$ and $t$. Indeed, note that if we let $u=\gcd(a,t)$, we have \[\frac{a}{u}\cdot \frac{a+t}{u}=\left(\frac{t}{u}\right)^2\cdot (t+2),\]with $\frac{a}{u}$, $\frac{a+t}{u}$ and $\frac{t}{u}$ being pairwise coprime integers. It thus follows from \[\left(\frac{t}{u}\right)^2\mid \frac{a}{u}\cdot \frac{a+t}{u}\]that $\frac{t}{u}\in \{-1,1\}$, or in other words $t\mid a$. We can thus let $a=k\cdot t$, for some integer $k$. Plugging this into $(*)$ yields \[k(k+1)=t+2\]\[\iff t=k^2+k-2=(k+2)(k-1),\]so $y+1=a=k(k+2)(k-1)$ and $x-1=a+t=(k+1)(k+2)(k-1)$. We thus have \[(x,y)=\Big((k+1)(k+2)(k-1)+1, k(k+2)(k-1)-1\Big)\]Note that all such pairs $(x,y)$ satisfy the equation $(*)$, which is equivalent to the problem statement. It just remains to check for which integers $k$ do we have $x,y\in \mathbb N$ and $x\ge y$. For $y$ to be positive we must have $k(k+2)(k-1)>0$, so $k\in \mathbb N_{>1}\cup \{-1\}$. This is actually sufficient. Indeed, we can easily check that all such $k$ give positive integer values to $x$ and $y$ with $x\ge y$. Finally, the set of all pairs $(x,y)\in \mathbb N_0^2$ which satisfy the problem statement are all pairs of the form (up to permutation) \[(x,y)=\Big((k+1)(k+2)(k-1)+1, k(k+2)(k-1)-1\Big),\]for some $k\in \mathbb N_{>1}\cup\{-1\}$. $\blacksquare$

WLOG $x=y+d$ with $d\ge 0$. Then we simplify to \[y^2+yd+7d^2-(d+1)^3=0.\]The discriminant of the resulting quadratic is $4d^3-15d^2+12d+4=(d-2)^2(4d+1)$ hence write $d=n^2+n$ for $n\ge 0$. The smaller root of the quadratic is positive if $d=0$, yielding $(1,1)$ as a solution. Otherwise we get $(x,y)=(n^3+2n^2-n-1,n^3+n^2-2n-1)$ and the permutation, with $n\ge 2$. $\blacksquare$

$50 and an icecream cake that Titu wrote this. Oh wait.... he did. WLOG $x \geq y$; then, sub $a = x-y$ and $b = x+y$, so that the equation becomes \begin{align*} \sqrt[3]{\frac{27(x-y)^2 + (x+y)^2}{4}} &= x-y+1 \\ \sqrt[3]{\frac{27a^2 + b^2}{4}} &= a+1 \\ 27a^2 + b^2 &= 4(a+1)^3 \\ b^2 &= 4a^3 - 15a^2 + 12a + 4 \\ b^2 &= (4a+1)(a-2)^2. \end{align*}So, $4a+1$ must be a perfect square; we must have $a = k(k+1)$ for some integer $k$; then, $b = (2k+1)(k+2)(k-1)$. In order for $a$ to be nonnegative and for $b$ to be positive, we must have $k \in \{ -1 \} \cup [2, \infty )$. Substituting back into the original variables, we have \[(x,y) = (k^3 + 2k^2 - k - 1, k^3 + k^2 -2k - 1)\]where $k$ is a positive integer in the range $k \in \{ -1 \} \cup [2, \infty )$. (Or, $x$ and $y$ can be swapped too.)

DISASTER DISASTER DISASTER DISASTER DISASTER DISASTER WLOG $x \ge y$. $x = y + n$ for $n$ a positive integer. We expand and observe the given is equivalent to \[ \sqrt[3]{y^2 + ny + 7n^2} = n + 1. \]Cubing, we have \[ y^2 + ny + 7n^2 - (n+1)^3 = 0. \]Now by the quadratic formula we must have \[ -27n^2 + 4(n+1)^3 = n^2 - 4(7n^2 - (n+1)^3)) = m^2 \]for a positive integer $m$. Then, observe that this is the cubic discriminant of the polynomial $x^3 - (n+1)x + n$, and after observing that this polynomial is equal to $(x-1)(x^2+x-n)$ we find that by cubic discriminant \[ (2-n)^2(1 + 4n) = m^2. \]Particularly, $1 + 4n$ is a perfect square, so $n = k(k+1)$ for some integer $k$. Now we have \[ y = \frac{-k(k+1) \pm (k + 2)(k-1)(2k + 1)}{2}. \]We take the positive root: \[ y = \frac{-k(k+1) + (k+2)(k-1)(2k+1)}{2} = \frac{2k^3 + 2k^2 - 4k + 2}{2} = k^3 + k^2 - 2k + 1, \]and we find that $x = k^3 + 2k^2 - k + 1$. Particularly either $k = -1$ or $k \ge 2$. Then we can swap x and y for other solutions, yay.