Find all natural numbers $n$ such that the sum of the squares of the positive divisors of $n$ is equal to $n^2+2n+37$.

Question

If $f(n)$ is the sum of the squares of the positive divisors of n,then find all natural numbers $n$ such that $f(n)=n^2+2n+37$. I tried it first by substituting few values of $n$ but that doesn't work I think. So, then I consider two cases where one is that $n$ is prime but it is not possible. Another is that $n$ is composite and here I got stuck. Any help would be appreciated.

Answer 1

Let us call the function $f$ as $\sigma_2$ since that's standard notation. Then, we have $\mathbb N\ni n\mapsto \sigma_2(n)=\sum_{d\mid n}d^2$. We start with a few lemmas and propositions some of which are easy enough so I will leave the proofs to you.

Lemma 1  $\sigma_2$ is multiplicative, that is, if $\gcd(a,b)=1$ then $\sigma_2(ab)=\sigma_2(a)\sigma_2(b)$.

Lemma 2  $n$ is prime $\iff$ $\sigma_2(n)=1+n^2$

Proof.  The only if part is simple. If $n$ is a prime, then it's only divisors are $1,n$ and hence $\sigma_2(n)=1^2+n^2=1+n^2$. For the other direction, let $p$ be the smallest prime factor of $n$. We write $n=mp$ and for the sake of contradiction, assume $m>1$. Clearly $m\ge p$. In fact the equality is strict because if $m=p$, then $1+p^4=\sigma_2(n)=\sigma_2(p^2)=1+p^2+p^4$ which is absurd. Thus $m>p$. Now, $$1+m^2p^2=\sigma_2(mp)\ge 1+m^2+p^2+m^2p^2\Rightarrow m^2+p^2\le 0$$ which is not possible and thus we have reached a contradiction, and hence $m=1$, completing the proof. $\square$

For ease of writing, let $h(n)=n^2+2n+37$.

Proposition 3  If $\sigma_2=h$, then $n\ne p^2$ for any prime $p$.

Proposition 4  Let $p,q$ be primes with $q>p$ and $\sigma_2=h$. Then $n=pq\Rightarrow q=p+6$.

Proof.  Simple calculation. $$p^2q^2+2pq+37=h(pq)=\sigma_2(pq)=1+p^2+q^2+p^2q^2$$ $$\Rightarrow (p-q)^2=36\Rightarrow q=p+6\ \ \ \ \ \ \square$$

Proposition 5  Let $p>3$ be a prime and $\sigma_2=h$. Then $n=p(p+6)\Rightarrow p+6$ is a prime.

Proof.  We note that $\gcd(p,p+6)=1$ because if not, then $p=2,3$ but $p>3$. Then $$\sigma_2(p(p+6))=\sigma_2(p)\sigma_2(p+6)=(1+p^2)\sigma_2(p+6)$$

Also

\begin{align*} \sigma_2(p(p+6))=h(p(p+6))&=p^2(p+6)^2+2p(p+6)+37\\&=p^2(p+6)^2+2p(p+6)+36+1\\&=p^2(p+6)^2+2p(p+6)+(p+6-p)^2+1\\&=p^2(p+6)^2+p^2+(p+6)^2+1\\&=(p^2+1)((p+6)^2+1) \end{align*}

Thus $\sigma_2(p+6)=1+(p+6)^2$ and hence by lemma 2, we are done. $\square$

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The preparations are complete. The trick in the proof of Lemma 2 is going to help again. Let $p$ be the smallest prime factor of $n$, and let $n=mp$. We observe that $m>p$ as $m=p$ does not work by proposition 3, and if $m<p$, then $p$ is not the smallest prime factor of $n$. Then since $\sigma_2=h$, $$m^2p^2+2mp+37=\sigma_2(mp)\ge 1+m^2+p^2+m^2p^2\Rightarrow (m-p)^2\le 36\Rightarrow m-p\le 6$$ and hence $$p+6\ge m>p$$

We note that the upper bound is tight by proposition 4. Then $n=p(p+k),\ 1\le k\le 6$.

Theorem 5  The only $n$ that works are $n=p(p+6),\ p\ge 3$ prime.

Remark.  The above theorem along with proposition 5 essentially says that if $\sigma_2(n)=h(n)$, then $n=pq$ where $(p,q)$ is a sexy prime pair.

Proof.  We can check that for $p=2$, this doesn't work. Also, for $p=3$, only $3(3+6)=27$ works (12 cases to check, not that difficult). Then let $p>3$.

Since primes are odd, $n=p(p+k),\ k=1,3,5$ doesn't work as then the smallest prime factor of $n$ is 2, and thus $p=2\not\ge 3$. Also, by proposition 5, $n=p(p+6)$ works with $p+6$ prime. Thus the remaining cases are $n=p(p+k),\ k=2,4$. Then we note that for both these cases $p+2,\ p+4$ have to be prime respectively, because if not, then some $p'$ prime divides $p+2$ (p+4 respectively), and then $p'<p$ which is a contradiction to the minimality of $p$.

Then by proposition 4, both these cases do not work, and hence we are done. $\square$

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An interesting observation is that $\sigma_2(n)=n^2+2n+(4k^2+1)$ holds only for $n=pq,\ q=p+2k$. The proof is essentially the same.