If n is odd we get contradition $ d_1^2+d_2^2+d_3^2+d_4^2$ - is even. Therefore $ d_1=1,d_2=2$. Because $ n>d_4^2$ n had $ k\ge 8$ divisors and $ d_{k+1-4}=\frac{d}{d_4}=d_4+\frac{d_3^2+5}{d_4}\le 3d_4-5.$ Therefore $ n\le d_4(3d_4-5).$ If $ m$ number of prime divisors equal 1 $ d_{k-3}$ is not integer, therefore $ m\ge 2$. If $ m=p_1^{k_1}p_2^{k_2}$. It is easy to chek, that for small prime $ p_1<p_2$ degree $ k_1\ge 2$. If $ p_2<p_1^2$ we get $ d_2=p_1,d_3=p_2,d_4=p_1^2|p_2^2+1\to p_1>2.$ It give contradition ($ 1+d_1^2+d_2^3+d_4^2$ is even). If $ p_2>p_1^2$ we get $ d_2=p_1,d_3=p_1^2,d_4=p_2|1+p_1^2+p_1^4=(p_1^2+1-p_1)(p_1^2+p_1+1).$ Therefore $ p_2=p_1^2+p_1+1$ and $ n=1+p_1^2+p_1^4+p_2^2=p_2(p_2+p_1^2-p_1+1)=2p_2(p_1^2+1)$ - contradition. Therefore $ m\ge 3$. If $ m>3$ we get contradition $ p_1p_2p_3p_4\ge n=d_1^2+d_2^2+d_3^2+d_4^2\le 1+p_1^2+p_2^2+p_3^2<3p_3^2$. If $ n=2^{k_1}p_2^{k_2}p_3^{k_3}$ by these whay we get $ k_3=1$. Chek by mod 4 give, that $ k_1=1$. Therefore $ n=2*p^kq$ p<q odd prime divisors. If $ q>2p$, then $ n=1+2^2+p^2+(2p)^2=5(p^2+1)\to p=5,q=13$ solution unique $ n=130$, because if $ p<q<2p$ from $ 2p^kq=5+p^2+q^2<5p^2$ had not solution.

See also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=22002