1

I don't have real experience in number theory or even a real reason to ask this question, but I was wondering about the relationship between $\frac xy$ and $\frac{x-1}{y-1}$. I originally assumed that if $x$ is divisible by $y$, then intuitively $x-1$ won't be divisible by $y-1$. This is wrong because $x=64$, $y=8$ gives $$\frac xy=\frac{64}{8}=8$$ $$\frac{x-1}{y-1}=\frac{63}{7}=9$$

My question is what relationship or information about natural numbers $x$ and $y$ can you find if $\frac xy$ and $\frac{x-1}{y-1}$ are integers?

Mittens
  • 46,352
MumboJumbo
  • 165
  • See AoPS. – Dietrich Burde Feb 23 '25 at 12:35
  • Note: The AoPS duplicate only addresses the case in which every divisor of $x$ has this property. Your example, with $63$ to line it up with the AoPS statement, doesn't...sure it works for $y=7$ but not for $y=9$. Unsurprisingly, the stronger version has very few (families of) solutions. Your version has a lot more. – lulu Feb 23 '25 at 12:48
  • 1
    Note that for your problem, we need only apply the Chinese Remainder Theorem to the congruences $n\equiv 0 \pmod m$ and $n\equiv 1\pmod {m-1}$ and remark that $\gcd(m,m-1)=1$. – lulu Feb 23 '25 at 12:59
  • All the square numbers (accompanied with their square roots) satisfy this property:

    $$\frac{x^2}{x} = x, \frac{x^2 - 1}{x - 1} = x + 1,$$

    and your example is one case of this.

     – Mathemagician314 Feb 23 '25 at 13:07
  • 2
    How about $$x=y^n$$ for natural $n$? – lab bhattacharjee Feb 23 '25 at 13:14
  • $x$ even, $y=2$. $x$ an odd multiple of three, $y=3$. – Gerry Myerson Feb 23 '25 at 23:22
  • By Easy CRT in the linked dupe: $$\begin{align}x&\equiv 0!!!\pmod{!y}\ x&\equiv 1!!!\pmod{!y!-!1}\end{align}!!\iff x\equiv y!!!\pmod{!y^2!-!y}\iff x = y+ n(y^2!-!y),\ n\in\Bbb Z$$ – Bill Dubuque Feb 24 '25 at 00:34

1 Answers1

1

In truth, it's a very broad church. Let's assign some labels to the results of those divisions, setting $x = ky$ and $(x - 1) = m(y - 1)$. By combining these equations, we get:

$$\begin{eqnarray} x - 1 & = & m(y - 1) \\ ky - 1 & = & my - m \\ (m - k)y & = & m - 1 \end{eqnarray}$$

So if we choose an $m$, we can pick one of the divisors of $m - 1$ to set as $m - k$, and that will give us our $x$ and $y$.

For every choice of $m$ we will always have the options of $k = 1$ and $k = m - 2$ (although for $m = 2$ these are the same thing). Setting $k = 1$ gives the trivial case where $x = y$, whereas setting $k = m - 2$ gives $y = m - 1$ and $x = (m - 1)^2$, where we then have $x - 1 = m^2 - 2m = m(m - 2) = m(y - 1)$ as required. For example, when $m = 9$ we get the $x = 64$, $y = 8$ solution you noted.

If $m - 1$ is composite, we have additional options. For example, sticking with $m = 9$ we could choose $k = 7$, giving $x = 28$ and $y = 4$. Or if $k = 5$ then we get $x = 10$, $y = 2$ (and in fact notice that any time $m$ is odd we will always have one solution of the form $x = m + 1$, $y = 2$ which we might also label as trivial).

ConMan
  • 27,579