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Let $n$ be a natural number, and let $s(n)$ denote the sum of all positive divisors of $n$. Show that for any $n>1$ the product $s(n-1)s(n)s(n+1)$ is always an even number.

I calculated the sum of the divisors(in the form a geometric Progression) and then tried proving. But, I was not able to comment on the divisors of $n-1$ and $n+1$. Please give some hints. Thanks.

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    $\begingroup$ @Moya It wouldn't be either $s(2)=1+2=3$ $\endgroup$ – user223391 Aug 30 '15 at 4:02
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    $\begingroup$ Not sure it helps, but the problem is equivalent to the following: "Given three consecutive positive integers, at least one has an even sum-of-divisors." $\endgroup$ – Semiclassical Aug 30 '15 at 4:27
  • $\begingroup$ @Moya It probably isn't the restricted divisor function since the first four values of that are $1,1,3,1$. $\endgroup$ – Erick Wong Aug 30 '15 at 4:27
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    $\begingroup$ @You-know-me Have you thought about the conditions under which $\sigma(n)$ is odd? Hint: try to show that when $n$ is odd, then $\sigma(n)$ is odd iff $n$ is a perfect square. What can you say about the case of $n$ even? $\endgroup$ – Erick Wong Aug 30 '15 at 4:29
  • $\begingroup$ I should mention that another notation for this function is $\sigma_1(n)$. $\sigma_0$ is the number of factors, $\sigma_1$ is the sum of the factors, $\sigma_2$ is the sum of the squares of the factors, $\sigma_3$ is the sum of cubes of the factors, etc. $\endgroup$ – Akiva Weinberger Aug 30 '15 at 5:13
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Express the numbers $n-1,n,n+1$ in their prime factorisations.

Let $x$ be any one of $n-1,n,n+1$ and express $$x=2^a\cdot3^b\cdot5^c\cdot\ldots \tag{1}$$

Then $$s(x)=(1+2+\ldots 2^a)\times(1+3+\ldots 3^b)\times(1+5+\ldots 5^c)\times\ldots \tag{2}$$

* Assume now that all of $s(n-1),s(n),s(n+1)$ are odd *

The first term is always odd, while the remaining terms are only odd if the exponents $b,c,\ldots$ are even, so $x$ must be a multiple of a power of 2 and an odd perfect square. As Erick Wong has commented this means that $x$ must be an odd perfect square if it is odd.

Now if $n-1,n+1$ are odd, they must both be odd perfect squares. But this is impossible as no two positive perfect squares differ by 2. Therefore, $n$ is odd, and hence an odd perfect square.

Express $n=\lambda^2$ where $\lambda=2k+1,\quad k\in\mathbb{Z^+}$. Then we have:

$$\begin{align} n-1 = \lambda^2-1 = (2k+1)^2 - 1 = 4k(k+1) \tag{3} \\ n = \lambda^2 = (2k+1)^2 = 4k^2+4k+1 \tag{4}\\ n+1 = \lambda^2+1 = (2k+1)^2 = 2(2k^2+2k+1) \tag{5} \end{align}$$

In (5), $2k^2+2k+1$ is odd, so must be an odd perfect square.

So $n+1$ is twice an odd perfect square.

In (3), one of $k,k+1$ must be odd. Also, $\gcd(k,k+1)=1$. So any odd prime factors in $k$ are not in $k+1$, and vice-versa.

Let $y$ be the even number in $\{k,k+1\}$ and $z$ be the other. Note that $y$ must be of the form $2^q\cdot r^2$ where $r$ is an odd number, and $z$ must be an odd perfect square, of the form $s^2$ where $s$ is odd. If $q$ is even, then $y$ is a perfect square, which is impossible because $y$ and $z$ would both then be positive perfect squares differing by 1. Therefore, we have shown that $q$ is odd, and so

$$4k(k+1)=4yz=4\cdot2^q\cdot r^2\cdot s^2=2\cdot2^{q+1}\cdot r^2\cdot s^2=2(2^{(q+1)/2}rs)^2$$

Since $q$ is odd, $(q+1)/2$ is an integer, and hence we can conclude that $n-1=4k(k+1)$ is twice a square number.

To recap, we now have that $n-1$ is twice a square number, and $n+1$ is twice an odd perfect square, with $n$ being odd. Therefore, $\frac{n-1}{2}$ and $\frac{n+1}{2}$ are both positive perfect squares. However, this is impossible because they differ by one. We have arrived at a fatal contradiction of the original assumption, so $s(n-1),s(n),s(n+1)$ cannot all be odd. $\blacksquare$

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