Much shorter: Get $a^2+3ab+3b^2-1\mid (a+b)^3$, and therefore, for any prime $p\mid a^2+3ab+3b^2-1$, we have $p\mid a+b$. Now, let $a^2+3ab+3b^2-1=\prod_{k=1}^L p_k^{a_k}$ with $a_k\geq 1$, $p_1<\cdots,p_L$ primes. Our goal is to show the existence of a $k$ with $a_k\geq 3$. Assume this is false, and that, $a_k\leq 2$, for every $k$. Now, notice that, $p_1^2\cdots p_k^2 \leq (a+b)^2$. This implies, $(a+b)^2 \geq a^2+3ab+3b^2-1$, yielding a contradiction.

Basically the same as above - The solution is motivated by the fact that \(((a^2+3ab+3b^2-1)a+(a+b^3)=(a+b)^3\), which yields \(a^2+3ab+3b^2-1\mid (a+b)^3\). Assume for contradiction \(\nu_p(a^2+3ab+3b^2-1)\le 2\) for all \(p\mid a^2+3ab+3b^2-1\). For any such \(p\), note that \(\nu_p(a+b)^3\ge 3\), so we can obtain that \(a^2+3ab+3b^2-1\mid (a+b)^2\), or \(a^2+3ab+3b^2-1\le (a+b)^2\iff ab+2b^2-1\le 0\), which is clearly false because \(a,b\ge 1\).

grupyorum wrote: Much shorter: Get $a^2+3ab+3b^2-1\mid (a+b)^3$, and therefore, for any prime $p\mid a^2+3ab+3b^2-1$, we have $p\mid a+b$. Now, let $a^2+3ab+3b^2-1=\prod_{k=1}^L p_k^{a_k}$ with $a_k\geq 1$, $p_1<\cdots,p_L$ primes. Our goal is to show the existence of a $k$ with $a_k\geq 3$. Assume this is false, and that, $a_k\leq 2$, for every $k$. Now, notice that, $p_1^2\cdots p_k^2 \leq (a+b)^2$. This implies, $(a+b)^2 \geq a^2+3ab+3b^2-1$, yielding a contradiction. I was this close to finishing the problem.... Now I feel horrible about myself.

fat Note $a(a^2 + 3ab + 3b^2 - 1) + (a + b^3) = (a + b)^3$ hence $a^2 + 3ab + 3b^2 - 1 \mid (a + b)^3$. Suppose the divisor is not divisible by any cube, then $v_p$ of it must be $\leq 2$ for all primes $p$ dividing it. Note that all primes $p$ dividing $a^2 + 3ab + 3b^2 - 1$ divide $(a + b)^3$ and thus $a + b$. This being said, for all such primes $p$, $v_p(a + b) \geq 1$ so $v_p((a + b)^2) \geq 2$ thus, in fact, $a^2 + 3ab + 3b^1 - 1 \mid (a + b)^2$ which is absurd.

Same as #3; posting for storage. Let $\ell=a^2+3ab+3b^2-1$ and note $$\ell\mid a+b^3-a(a^2+3ab+3b^2-1)=(a+b)^3.$$Suppose FTSOC that $\ell$ is not divisible by cubes of primes. Letting $a+b=\prod p_i^{\alpha_i}$ we see $(a+b)^3=\prod p_i^{3\alpha_i}.$ On the other hand, $\ell=\prod p_i^{\beta_i}$ where $\beta_i\le 2$ because it is cube-free. Hence, $\alpha_i+\beta_i\le 3\alpha_i$ and $\ell(a+b)\mid (a+b)^3,$ or $a^2+3ab+3b^2-1\mid a^2+2ab+b^2,$ which is absurd. $\square$

Suppose that $a^2+3ab+3b^2-1$ is not divisible by the cube of an integer greater than $1$. Consider prime $p$ dividing $a^2+3ab+3b^2-1$, then we obviously have $a \equiv -b^3 \pmod p$. This gives us: \begin{align*} 0 \equiv a^2+3ab+3b^2-1 \equiv \\ b^6-3b^4+3b^2-1 \equiv \\ \equiv (b^2-1)^3 \pmod p \end{align*} This means that either $b \equiv 1 \pmod p \implies a \equiv -1 \pmod p$ or $b \equiv - 1 \pmod p \implies a \equiv 1 \pmod p$. In both cases we have $a+b \equiv 0 \pmod p$. Since $\nu_p(a^2+3ab+3b^2-1) \le 2$, we conclude that: $$ a^2+3ab+3b^2 -1 \mid (a+b)^2 $$This means that: $$ a^2+2ab+b^2 \ge a^2+3ab+3b^2-1 \implies 0 \ge 2ab+b^2-1 $$which is an obvious contradiction since $a,b$ are positive integers.

This is literally the same problem as ISL 2021 N1.

Solved with GoodMorning. Let $S = a^2 + 3ab + 3b^2 - 1$. We have $S \mid a + b^3 + a \cdot S = (a+b)^3$. Since $S > (a+b)^2$, it can't be cubefree, done.

Assume, FTSOC, it's not. We have $a^2 + 3ab + 3b^2 - 1 \mid a + b^3 + a^3 + 3a^2b + 3ab^2 - a \implies a^2 + 3ab + 3b^2 - 1 \mid (a + b)^3$. Observe that $p \mid a^2 + 3ab + 3b^2 - 1 \implies p \mid (a + b)$, and $v_p(a^2 + 3ab + 3b^2 - 1) \leq 2 \leq v_p((a + b)^2)$, so $a^2 + 3ab + 3b^2 - 1 \mid (a + b)^2$, a clear contradiction by looking at size, and we're done.