# prime factorizations/ sum of squares of divisors.

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.

• Oh, and $p+6$ must also be prime for $\sigma_2(p(p+6))=n^2+2n+37$. Oct 4, 2022 at 12:37
• Yes, but this is the exact opposite of hte problem statement :P
– user1102993
Oct 4, 2022 at 13:09
• Seems that $27$ is the only such solution Oct 4, 2022 at 13:23
• not sure if this helps but $n^2+2n+37= (n+1)^2 + 6^2$. Oct 4, 2022 at 14:13
• In general, please avoid editing a question in a way which makes an existing correct answer incorrect. In this case, the change isn't a big deal: the existing answer led to a simple description of all possibilities for the modified question. Oct 4, 2022 at 23:35

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 $$\sigma_2(n)=n^2+1; it cannot be a squared prime for similar reasons. Thus $$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 $$n=27$$.

• Thank you so much! However, both $p$ and $p+6$ must be prime to be a prime factorization of n. I believe this was not mentioned properly, but here it is. With your example, we can find a counterexample with $27$, which is $3*9$ but $9$ is not prime.
– user1102993
Oct 4, 2022 at 22:59
• @godlification it is fixed now... Oct 4, 2022 at 23:50

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 $$n=27$$.