Ireland

First Solution: The key to this solution is the identity $$a^3+b^3-1+3ab=(a+b-1)(a^2+b^2-ab+a+b+1).$$ Even though this can be verified readily, we explain how to find it. We start with a polynomial of degree 3 which has $a$, $b$ and an auxiliary number $c$ as its roots $f(x)=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$. We obtain the identity $$0=f(a)+f(b)+f(c)=a^3+b^3+c^3-3abc-(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$ Substituting $c=-1$ gives the identity mentioned above.

The original equation $a^3+b^3+3ab=53$ is equivalent to the equation $$(a+b-1)(a^2+b^2-ab+a+b+1)=52.$$ Because we are looking for integer solutions, the two numbers $u=a+b-1$ and $v=a^2+b^2-ab+a+b+1=(a+b{)}^2+(a+b-1)-3ab+2=(u+1{)}^2+u+2-3ab=(u+1)(u+2)+1-3ab$ have to be divisors of $52=2\cdot 2\cdot 13$. The integer divisors of $52$ are $\pm 1,\pm 2,\pm 4,\pm 13,\pm 26,\pm 52$. In particular, $u\equiv \pm 1(\mathrm{mod}3)$. We have $a+b=u+1$ and $3ab=(u+1)(u+2)+1-v$. The last identity and $u\equiv \pm 1(\mathrm{mod}3)$ imply $v\equiv 1(\mathrm{mod}3)$. Hence, the following table covers all possibilities for $u$ and $v$.

u	v	a+b	3ab	ab
1	52	2	-45	-15
-2	-26	-1	27	9
4	13	5	18	6
13	4	14	207	69
-26	-2	-25	603	201
52	1	53	2862	954

From $x=a+b$ and $y=ab$ we obtain $(a,b)$ by solving the quadratic equation $a^2-ax+y=0$ and using $b=x-a$. For $a$ to be an integer, we need therefore that the discriminant of the quadratic, $x^2-4y$, is a perfect square. Only the first and third row of the table give $x^2-4y\ge 0$. Therefore, the only possible solutions are $(5,-3),(-3,5),(3,2),(2,3)$. Substituting them in the original equation shows that they are indeed solutions.

Second Solution: Because $53$ is not a perfect cube, neither $a$ nor $b$ could be zero. Without loss of generality, we may assume $a\le b$. We distinguish the following three cases.

$0<a\le b$: In this case, $a^3+b^3+3ab\ge 2a^3+3a^2=a^2(2a+3)$, hence $a^2(2a+3)\le 53$. This implies $a\le 2$. If $a=1$, we get $b^3+3b=b(b^2+3)=52=2\cdot 2\cdot 13$. As $b<b^2+3$, the only possibilities for $b$ are $1,2,4$ but none of them leads to a solution. If $a=2$, we have to solve $b^3+6b=b(b^2+6)=45=5\cdot 3\cdot 3$. As $b<b^2+6$, $b$ can only be $1,3$ or $5$. Only $b=3$ is a solution.

$a\le b<0$: In this case, $a^3+b^3+3ab=a(a^2+3b)+b^3=53>0$. Because $b^3<0$ and $a<0$ this implies $a^2+3b<0$. Hence, $a^2<3|b|$. On the other hand, $|a|\ge |b|$ and so $b^2\le a^2<3|b|$ which implies $|b|<3$. If $b=-1$, we get $a^3-3a=a(a^2-3)=54>0$. As $a<0$ this implies $a^2<3$, i.e. $a=-1$ which does not give a solution. If $b=-2$, we get $a^3-6a=a(a^2-6)=61$. As before, this implies $a^2<6$, i.e. $a=-1$ or $a=-2$. But both values do not solve the equation.

a < 0 < b: In this case, $3ab<0$ and so $a^3+b^3>0$ which implies $b>|a|$. Let $c=-a>0$ and $b=c+k$ with $k\ge 1$. The given equation becomes $$\begin{array}{rl}{b}^3-c^3-3bc& =53\\ (c+k{)}^3-c^3-3c(c+k)& =53\\ 3c^2(k-1)+3ck(k-1)+k^3& =53.\end{array}$$ The left hand side is at least $3(k-1)+3k(k-1)+k^3=k^2(k+3)-3$, hence we need to have $k^2(k+3)\le 56$ which implies $k\le 3$. We cannot have $k=1$, because $53$ is not a perfect cube. If $k=2$ we need to solve $3c^2+6c+8=53$, or equivalently, $c(c+2)=15$ with $c=3$ as its only positive solution. This leads to $(a,b)=(-3,5)$. Finally, if $k=3$ we need to have $6c^2+18c+27=53$ which has no solution as $53$ is not divisible by $3$.

Therefore, the only solutions to the original equation are $(5,-3),(-3,5),(3,2)$ and $(2,3)$.