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Notation: For an integer $k\ge1$, let $D(k)$ denote the sum of all positive integer divisors of $k$ that are strictly smaller than $k$. For example $D(9)=1+3=4$ and $D(33)=1+3+11=15$.


A man picked two odd positive integers $m\ge3$ and $n\ge3$. He computed the sums of divisors for $m$ and $n$, and multiplied them together. Then he suddenly realized that $$ D(m)\cdot D(n) ~=~ (m-1)(n-1)$$

Question: Is this indeed possible, or has the man made a calculation error?

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    $\begingroup$ I think that in the case of sum of prime factors or sum of distinct prime factors the answer is definitely no by simple magnitude limits. However, the sum of divisors can be large (for an abundant number) in which case the proof is more difficult, so that's probably the intended case. By enumeration, no solution exists for $n$ or $m$ less than 10k in any of the three cases. $\endgroup$ – 2012rcampion Oct 23 '15 at 7:03
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    $\begingroup$ Also note that we can simplify the condition to just $(m-1)(n-1) = x y$. $\endgroup$ – 2012rcampion Oct 23 '15 at 7:10
  • $\begingroup$ The smallest value $m$ so that $x\geq m-1$ is $m=945=3^3 \times 5 \times7$, which gives $x=975$. $\endgroup$ – xnor Oct 23 '15 at 8:40
  • $\begingroup$ Unless the poser has recently solved an open problem, the man must not have made a mistake in his calculation. Note that if $x$ is odd with $D(x)=x-1$, then $m=n=x$ solves the problem. However, such an $x$ is called "almost perfect" since $\sigma(x)=2x-1$ and the existence of odd "almost perfect" numbers is unknown. See mathworld.wolfram.com/AlmostPerfectNumber.html $\endgroup$ – DaveBlackston Apr 7 '16 at 22:41
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(Not a solution, just some partial progress:)

Lemma 1. If $m$ is odd and $D(m)$ is even, then $m$ is a perfect square.

Proof: Let $\prod_{i=1}^kp_i^{a_i}$ be the prime factorization of $m$. Then the sum of all divisors of $m$ (with $m$ itself included) is $$ D(m)+m ~=~ \prod_{i=1}^k (1+p_i+p_i^2+\cdots+p_i^{a_i})$$ If one of the exponents $a_i$ is odd, then the corresponding factor $1+p_i+p_i^2+\cdots+p_i^{a_i}$ in the right hand side has an even number of odd terms, and hence is even. Then the right hand side is even, which is in contradiction to $m$ odd and $D(m)$ even.

This contradiction implies that all exponents $a_i$ must be even, so that $m$ is a perfect square. Q.e.d.

Lemma 2. If that man's calculation is indeed correct, then at least one of the numbers $m$ and $n$ must be an odd perfect square.

Proof: By the problem statement both numbers $m$ and $n$ are odd. Suppose for the sake of contradiction that neither of them is a square. Then by Lemma 1, the numbers $D(m)$ and $D(n)$ both are odd, too. But then the left hand side $D(m)D(n)−1$ in the given equation is even, whereas its right hand side $mn−(m+n)$ is odd. Contradiction. Q.e.d.

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  • $\begingroup$ Tried to build on this, didn't get very far. For m a perfect square, D(m) must be < m, since all factors of m are less than or equal to the square root of m, so D(m) is the sum of less than sqrt(m) numbers each less than or equal to sqrt(m). So if m is a perfect square, n isn't. So D(n) is odd, and D(m) must be a multiple of 4. $\endgroup$ – histocrat Oct 23 '15 at 15:00
  • $\begingroup$ @histocrat: I also did not manage to come much further. Perhaps it is a dead end. On the other hand, the lemma exploits the oddness of $m$ and $n$ so nicely that there should be some connection to the solution. $\endgroup$ – Gamow Oct 23 '15 at 15:02

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