31st Hellenic Mathematical Olympiad

Let $p,q\in \mathbb{Z}$, $q\ne 0$, with $(p,q)=1$ such that $$A=\frac{8n-25}{n+5}={\left(\frac{p}{q}\right)}^3.(1)$$ Then $(p^3,q^3)=1$, while from relation (1) we get: $$q^3(8n-25)=p^3(n+5),(2)$$ from which we conclude that $$p^3\mid (8n-25)\text{ and }{q}^3\mid (n+5)\Rightarrow \exists k\in \mathbb{Z}:8n-25=k{p}^3\text{ and }n+5=k{q}^3.(3)$$ In fact, if $8n-25=k{p}^3$, $k\in \mathbb{Z}$, then from (2) we find $n+5=k{q}^3$ and so $$8(n+5)-(8n-25)=k(8p^3-q^3)\Rightarrow k(2q-p)(4q^2+2qp+p^2)=65$$ Therefore, the numbers $k,2q-p$ and $4q^2+2qp+p^2$ are divisors of 65. We observe that $4q^2+2pq+p^2=3q^2+(p+q{)}^2\equiv 1(\mathrm{mod}3)$. Moreover, we have $4q^2+2pq+p^2=3q^2+(p+q{)}^2\ge 3$, and so, since $4q^2+2pq+p^2$ is divisor of 65 its unique value is $$4q^2+2pq+p^2=13\iff 3q^2+(p+q{)}^2=13,$$ leading to the following cases: $4q^2+2qp+p^2=13$, $k=\pm 1$, $2q-p=\pm 5$. Then: $p=2q\mp 5$ and $4q^2+2q(2q\mp 5)+(2q\mp 5{)}^2=13\iff p=2q\mp 5$, $2q^2\mp 5q+2=0$ $\iff p=2q\mp 5$, $q=\pm 2\iff p=\mp 1$, $q=\pm 2$. Then for both cases we have: $${\left(\frac{p}{q}\right)}^3=-\frac{1}{8}\Rightarrow \frac{8n-25}{n+5}=-\frac{1}{8}\iff 8(8n-25)=-(n+5)\iff n=3.$$

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Alternative solution.

As in the first solution we obtain relation (2) which we solve with respect to $n$, to receive: $$n=\frac{5(5q^3+p^3)}{8q^3-p^3}.(4)$$ Let $d=(5q^3+p^3,8q^3-p^3)$. Then $d\mid (5q^3+p^3+8q^3-p^3=13q^3)$ and since $(p,q)=1$ it follows that $d\mid 13$. Therefore $$\frac{8q^3-p^3}{d}\mid 5\Rightarrow (8q^3-p^3)\mid 5\cdot 13\Rightarrow (2q-p)(4q^2+2pq+p^2)\mid 65.$$ As in the first solution we find that $$4q^2+2pq+p^2=13\iff 3q^2+(p+q{)}^2=13.(5)$$ Now from relation (5) it follows that $|q|\le 2$ and so we have the cases: If $|q|\in \{0,1\}$, then $(p+q{)}^2\in \{13,10\}$, impossible for $p,q\in \mathbb{Z}$. * If $|q|=2$, then $(p+q{)}^2=1$, from which we receive the solutions: $$(p,q)=(-1,2)\text{ or }(p,q)=(1,-2)\text{ or }(p,q)=(-3,2)\text{ or }(p,q)=(3,-2).$$ Since $(2q-p)|65$, acceptable are the solutions: $(p,q)=(-1,2)$ and $(p,q)=(1,-2)$, from which we find $n=3$.