Find all positive integers $n$ such that the sum of the squares of the divisors of $n$ is equal to $n^2+2n+37$, and in which $n$ is not of the form $p(p+6)$ where p and p+6 are prime numbers.
I tried to simplify this by using the prime factorization of n in the equation. However, this did not help. I also noticed that this is true for all $n=p(p+6)$, but this is the opposite of what the question asked.
On
Since $n$ and $1$ are factors of $n$ the condition becomes $$\sum f_i^2=2n+6^2$$ where each $f_i$ is a divisor of $n$ distinct of $n$ and $1$. So if $$n=p_1^{n_1}p_2^{n_2}\cdots p_r^{n_r}$$ we should have $$(p_1^{2n_1}+p_2^{2n_2}+\cdots+ p_r^{2n_r})+H=2p_1^{n_1}p_2^{n_2}\cdots p_r^{n_r}+6^2$$ where $H$ is in general an integer much greater than $6^2$ and because of, in general for positive integers,$$(p_1^{2n_1}+p_2^{2n_2}+\cdots+ p_r^{2n_r})\gt 2p_1^{n_1}p_2^{n_2}\cdots p_r^{n_r}$$ we would have $H+\Delta=6^2$ where $(p_1^{2n_1}+p_2^{2n_2}+\cdots+ p_r^{2n_r})+\Delta= 2p_1^{n_1}p_2^{n_2}\cdots p_r^{n_r}$.Then we can see that $n$ has not several prime factors (which one can show more formally, of course).
$(1)$ Case where $n=p^k$
►$p\ne2$ because if not then $2^2(1+2^2+\cdots+2^{2k-4})=2^{k+1}+6^2\Rightarrow2^2+\cdots+2^{k-4}=2^{k-1}+8$ which implies $2^2+\cdots+2^{k-6}=2^{k-3}+1$, absurde if $k\ge6$ and a fortiori if not (because fraction= integer).
►►Let $n=p^k$ so $p^2+p^4+p^6+\cdots+p^{2(k-1)}=2p^k+6^2$. So if $p$ is odd then $k$ must be odd because if not we would have odd=even. If $k=1$ then $0=2p+36$ absurd, thus $k$ is odd greater or equal to $3$. Find this way, direct calculation gives the only example $n=27$.
$(2)$ Case of two prime factors.
If $n=pq$ then $p^2+q^2=2pq+6^2$ so $(p-q)^2=6$ and $n=q(q+6)$ wich is not admissible by the problem. When $p$ or $q$ have an exponent greater than $1$ we apply the reasoning above showing that it is not possible. And nor for three or more prime factors.
Consequently the only solution is $n=27$.
First we show that $n$ cannot be even: if so, $\sigma_2(n)\ge n^2+n^2/4+4+1$, which is strictly greater than $n^2+2n+37$ for $n>16$. Checking even $n\le16$ shows no solutions.
$n$ cannot be prime since then $\sigma_2(n)=n^2+1<n^2+2n+37$; it cannot be a squared prime for similar reasons. Thus $n=(k+d)(k-d)$ for $k>d>0$ with $k-d$ odd and at least $3$ and $$\sigma_2(n)\ge n^2+(k+d)^2+(k-d)^2+1=n^2+2k^2+2d^2+1=n^2+2n+4d^2+1$$ If $d\ge4$ there can be no equality with $n^2+2n+37$. It is easy to show that with $d\le3$, there is only one choice for $d$, so the only nontrivial divisors of $n$ are $k+d$ and $k-d$. This fact implies $k-d$ is prime and $\sigma_2(n)=n^2+2n+4d^2+1$ after all, so $d=3$ and $n=(k-3)(k-3+6)$ with $k-3$ prime.
Either $k+3$ is also prime or it is $(k-3)^2$, leading to the sole exception of $n=27$.