I recently encountered the following problem:

A natural number $n$ is called correct if and only if the sum of the squares of $n$'s divisors equals $(n+3)^2$. Prove that no numbers of the form $n=p^m$, where $p$ is a prime and $m\in\Bbb{N}$ (prime powers) can be correct. (from the N. N. Mihaileanu Mathematics Contest – 8th Grade 2017, Romania)

First of all, I looked at the sum of the divisor squares:

$$S=\sum_{d\in D_{p^m}^+}d^2$$

Where $D_{p^m}^+$ is the set of $p^m$'s positive divisors. However, since $p^m$ is a prime power, this set can also be expressed as:

$$D_{p^m}^+=\{0\le k\le m\mid p^k\}$$

So it follows that our sum can simply be expressed as:

$$S=\sum_{k=0}^m (p^k)^2=\sum_{k=0}^m p^{2k}$$

Now, I tried to continue on with the proof using Reductio ad absurdum, by assuming that $S$ can equal $(n+3)^2$:

$$\sum_{k=0}^m p^{2k}=(p^m+3)^2$$

Subtracting $p^{2m}$ from both sides,

$$\sum_{k=0}^{m-1} p^{2k}=6p^m+9$$

Then, I tried evaluating the sum of that series, obtaining:


However, I am stuck here. I tried my best to avoid splitting in cases, but am currently unable to finish my proof. Any hints on what I should do next (preferably, without having to consider cases separately, but that works too)?

  • 1
    $\begingroup$ Why avoid modular arithmetic? $\endgroup$ – Kenny Lau May 22 '18 at 10:39
  • $\begingroup$ @KennyLau Mainly because I'd have to consider $2$ separately, at least $\mod p$. I realised I don't want to avoid MA, but rather taking things separately. I am fine with that if it's really needed, but I prefer not to deal with it :) $\endgroup$ – Mr. Xcoder May 22 '18 at 10:43
  • $\begingroup$ From that sum you get $6p^m-6p^m+9p^2-2p^2m=8$ so $modp$ this means $8=0modp$ which implies $p=2$. So you only need to examine that case. $\endgroup$ – Μάρκος Καραμέρης May 22 '18 at 11:10

Letting $x=p^m$, the key equation becomes $6x+9=(x^2-1)/(p^2-1)$, or


for which the quadratic formula tells us


This implies the discriminant is a perfect square, which implies


is a square. But the only two squares that differ by $5$ are $9$ and $4$, and

$$6p^2-3=3\implies p=\pm1$$

  • $\begingroup$ Absolutely beautiful argument, only using the quadratic formula and some clever tricks. Awesome! $\endgroup$ – Mr. Xcoder May 22 '18 at 12:37
  • 1
    $\begingroup$ This is arguably the best answer, imho. Unfortunately I have reached my daily voting limit, but if I didn't, $(+1)$ $\endgroup$ – Mr Pie May 25 '18 at 9:54

I think the simplest proof is a $\mod p$ argument (as suggested in one of the comments to the original question):

If $n=p^m$ then all of the divisors of $n$ apart from 1 are multiples of $p$, so

$\sum_{d|n} d^2 \equiv 1 \mod p$

and $(n+3)^2 = (p^m+3)^2 \equiv 9 \mod p$

If $1 \equiv 9 \mod p$ then $p=2$ and $p^m \in \{2,4,8,16\}$ - after 16, $\sum d^2 > (2^m+3)^2$. But we can eliminate $\{2,4,8,16\}$ by direct calculation. So

$\sum_{d|n} d^2 \ne (n+3)^2 \quad \forall n=p^m$

  • $\begingroup$ Why $p^m\in\{2,4,8\}$? $\endgroup$ – Mr. Xcoder May 22 '18 at 12:26
  • $\begingroup$ Should be $\{2,4,8,16\}$ - I have corrected my answer. $\endgroup$ – gandalf61 May 22 '18 at 13:17

I'll continue the argument from $$\sum_{k=0}^{m-1} p^{2k}=6p^m+9\tag{1}\label{eq1}$$

Firstly, assume $m=1$ or $m=2$. Then $\eqref{eq1}$ becomes $1=6p+9$ or $1+p^2=6p^2+9$, which is impossible. Thus $m\ge3$.

i) $p=2$. Then since $2^{2(m-1)}<6\times2^m+9<8\times2^m=2^{m+3}$, $m<5$. Manually checking the cases $m=3$ and $m=4$ gives no solutions.

ii) $p=3$, then $3^{2(m-1)}<6\times3^m+9<9\times3^m=3^{m+2}$, $m<4$. Manually checking the case $m=3$ gives no solution.

iii) $p=5$, then $5^{2(m-1)}<6\times5^m+9<25\times5^m=5^{m+2}$, $m<4$. Manually checking the case $m=3$ gives no solution.

iv) $p\ge7$, then $p^{2(m-1)}<6p^m+9<7p^m\le p^{m+1}$, $m<3$.

By i) to iv), we can conclude that no primes can be correct number.

Also, one can replace the arguments of ii) and iii) to modular arithmetic.


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