Can we write a formula for the least common multiple of the two rational numbers $$ \frac{n^2}{n^2 - 1} \mbox{ and } \frac{(n+1)^2}{(n+1)^2 - 1} $$ for $n > 1$ a natural number? Numerical investigations hint that the LCM diverges as $n \rightarrow \infty$: how can one prove (or disprove) this?
2026-02-24 05:37:27.1771911447
Least common multiple of $\frac{n^2}{n^2 - 1}$ and $\frac{(n+1)^2}{ (n+1)^2 - 1 }$
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The LCM of two rational numbers in lowest terms is the LCM of the numerators divided by the GCD of the denominators. It's pretty clear that both of these fractions are in lowest terms by checking that both fractions have a difference of $1$ between their numerators and denominators, so their GCD must be $1$.
$n^2$ and $(n+1)^2$ have no common factors, so their LCM is just their product, or $n^2(n+1)^2$.
$n^2-1=(n+1)(n-1)$ and $(n+1)^2-1=(n+1+1)(n+1-1)=(n+2)n$. The differences between the divisors of the former and of the latter are $1, 3$, so the GCD could be $1$ or $3$. However, both of these numbers are constants, so we'll just refer to the GCD as $C$.
Thus, we have the LCM of these two rational numbers to be $\frac{n^2(n+1)^2}{C}$, which indeed diverges as $n \to \infty$.