For any integer $n$, prove that LCM $[9n+ 8, 6n+ 5] = 54n^2+ 93n+ 40$.

Question

I am having trouble where to start on this proof. I see that when you multiply $9n+8$ and $6n+5$ you get $54n^2+93n+40$. I think you might have to do something with relative primes. Please help me prove this. Thanks!

Answer 1

Hints: As you have observed, the product of the two numbers is the RHS of your desired result. Therefore, if you can show that the two numbers are coprime (have gcd 1), your result will follow (because if $a$ and $b$ are coprime, then $\mathrm{lcm}(a,b) = ab$, for any positive integers $a,b$).

So try and show that $9n+8$ and $6n + 5$ are coprime (one possibility: Bezout's identity).

Answer 2

For any two integers $c,d$ you have

• $gcd(c,d)\cdot lmc(c,d) = |cd|$

So, you need to show that $gcd(c,d) = 1$.

This you can do using Euclid's algorithm as follows: \begin{eqnarray*} 9n + 8 & = & 1\cdot (\color{blue}{6n + 5}) + \color{green}{3n+3} \\ \color{blue}{6n + 5} & = & 2\cdot (\color{green}{3n+3}) - \color{red}{1} \\ \end{eqnarray*}

Now, note that

• $gcd(c,d) = 1 \Leftrightarrow$ there are two integers $x,y$ such that $xc+yd = 1$.

The result of the algorithm shows that we have found $k,l$ such that $k\cdot (\color{blue}{6n + 5}) + l(\color{green}{3n+3}) = \color{red}{-1}$. Multiplying this equation by $-1$ gives the required result.

Answer 3

Recall $\ {\rm lcm}(a,b) = ab\iff \gcd(a,b) = 1.$ But it's easy to show the gcd $=1$ here:

Hint $ $ Eliminate $n,\,$ i.e. $\ 2(9n\!+\!8)-3(6n\!+\!5) = 1\ $ so $\,d\mid 9n\!+\!8,\,6n\!+\!5\,\Rightarrow\,d\mid 1$