An equation about divisor function

Question

We define the 2th divisor function as follows: $\sigma_2(n)=\sum_{d\mid n}d^2$?

Is there infinite positive integer $n$ such that $\sigma_2(n)=(n+3)^2$? If there is, how can we find the least $n$?

Answer 1

Sketchwe list the divisors of $n$:$$1=d_1<d_2<\dots<d_k=n$$ then \begin{aligned} &1+d_2^w+\cdots+d_{k-1}^2+n^2=(n+3)^2\\ &\iff d_1^2+\cdots+d_{k-1}^2=6n+8 \end{aligned} now we should estimate the range of$k$.

1. If $k\ge8$, \begin{aligned} &\quad d_1^2+\cdots+d_{k-1}^2\\&\ge (d_2-d_{k-1})^2+2d_2d_{k-1}+(d_3-d_{k-2})^2+2d_3d_{k-2}+(d_4-d_{k-3})^2+2d_4d_{k-3}\\&\ge 6n+8 \end{aligned} A contradiction!
2. If $3\le k\le 7$,a little difficult
3. $k=2$,let $n=pq$($p,q$ are different prime),then $$p^2+q^2=6pq+8$$ $$\iff (p-3q)^2=8q^2+8$$ $$\iff q^2-2\left(\frac{p-3q}{4}\right)^2=-1$$ it is similar to Pell equation! $$x^2+2y^2=-1$$ let $$x_n-\sqrt2 y_n=(1+\sqrt2)^{2n-1}$$ luckily,$$x_2=5,y_2=7\implies p=41,q=7\implies n=287$$ $$x_3=41,y_3=29\implies p=239,q=41\implies n=9799$$