Brazilian Math Olympiad

Let $n=p_1^{{\alpha }_1}\cdots p_k^{{\alpha }_k}$. Then $8=({\alpha }_1+1)\cdots ({\alpha }_k+1)$, and $3240=\frac{p_1^{{\alpha }_1+1}-1}{p_1-1}\cdots \frac{p_k^{{\alpha }_k+1}-1}{p_k-1}$. Hence there are three cases: (a) $8={\alpha }_1+1⟹n=p^7$. (b) $8=({\alpha }_1+1)({\alpha }_2+1)⟹n=p_1{p}_2^3$. (c) $8=({\alpha }_1+1)({\alpha }_2+1)({\alpha }_3+1)⟹n=p_1{p}_2{p}_3$. Then we check $\sigma (n)$: (a) $1+p^2+\cdots +p^7=3240$. We have $2<p<5$, so $p=3$. But substituting yields no solution. (b) $(p_1+1)(p_2^3+p_2^2+p_2+1)=3240⟺(p_1+1)(p_2+1)(p_2^2+1)=3240$. First note that if $q$ is an odd prime then, by Fermat's theorem, $p_2^2+1\equiv 0(\mathrm{mod}q)⟹1\equiv p_2^{q-1}\equiv (-1{)}^{\frac{q-1}{2}}(\mathrm{mod}q)⟹q\equiv 1(\mathrm{mod}q)$. Since $3240=2^3\cdot 3^4\cdot 5$, the only primes that can divide $p_2^2+1$ are $2$ and $5$. This leaves the possibilities $p_2=2$ and $p_2=3$, none of which yield a solution.

(c) $(p_1+1)(p_2+1)(p_3+1)=3240$. Let's consider some cases. (c.1) One of the primes $p_i$ is $2$. Suppose that $p_1=2$. Then $(p_2+1)(p_3+1)=1080⟺(\frac{p_2+1}{2})(\frac{p_3+1}{2})=270$. Let $x=\frac{p_2+1}{2}$ and $y=\frac{p_3+1}{2}$. Then $xy=270$ is fixed and we want to minimize $2p_2{p}_3=2(2x-1)(2y-1)=8\cdot 270-4(x+y)+2$, that is, we want to maximize $x+y$. This happens when $|x-y|$ is maximum. Since $p_2$ and $p_3$ are primes, the optimal values for $x$ and $y$ are $x=2$ and $y=135$, that is, $p_2=3$ and $p_3=269$, leading to the minimal solution $n=1614$. (c.2) All primes $p_i$ are odd. Then $(\frac{p_1+1}{2})(\frac{p_2+1}{2})(\frac{p_3+1}{2})=405=3^3\cdot 5$. Then one prime, say $p_1$, is equal to $2\cdot 3^k\cdot 5-1=10\cdot 3^k-1$; and since $\frac{p_i+1}{2}$ cannot be equal to $1$, then they must have at least one factor $3$ for $i=2,3$; there are only three factors $3$, so $k=0$ or $k=1$. $k=0$ is not possible; $k=1$ yields $p_1=29$ and $(\frac{p_2+1}{2})(\frac{p_3+1}{2})=27⟹p_2=5$ and $p_3=17$, leading to $n=2465$. Hence the smallest value for $n$ is $1614$.