<s>[INST] Solving integer equation $a^3+b^3=a^2+72ab+b^2$ 
Find all pairs of positive integers $(a;b)$ that satisfy
  $$a^3+b^3=a^2+72ab+b^2.$$

I have already solved this by letting $S=a+b$, $P=ab$. Then I have $S^3-S^2=(3S+70)P$, which will result in $3S+70$ dividing $5110S$, or $357700$ is divisible by $3S+70$. The result is $(a;b)\in\left\{(1;9),(9;1),(37;37)\right\}$.
However, this strategy is extremely complicated and time-consuming, by listing all divisors of $357700$. I do not know any other way to access this problem. Many thanks.
 [/INST] Let $a=x+y, b=x-y$. Then:
$$a^3+b^3=a^2+72ab+b^2 \Rightarrow \\
(x+y)^3+(x-y)^3=(x+y)^2+72(x+y)(x-y)+(x-y)^2 \Rightarrow \\
2x^3+6xy^2=2x^2+72x^2 -72y^2+2y^2 \Rightarrow \\
y^2=\frac{(37-x)x^2}{35+3x}\ge 0.$$
Note that:
$$x=\frac{a+b}{2}>0 \ \ \text{and} \ \  37-x\ge0  \Rightarrow 0<x\le 37.$$
One can quickly check (if not further analysis) all numbers to find: $(x,y)=(5,4),(37,0)$. 
Hence: 
$$(a,b)=(9,1),(37,37) \ \ \text{and} \ \ (1,9) \ \text{(due to symmetry)}.$$
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