Lowest common multiple proof For any positive integer $n$, show that $lcm (9n+8, 6n+5)=54n^2+93n+40$
I know this is true if you expand $(9n+8)(6n+5)$ but I know there is more to this proof than just that I'm just not sure where to start.
 A: You've done half of it.  $(6n+5)*(9n+8)$ must be a common multiple ... because it is a multiple of both $9n+8$ and $6n+5$.
But is it the least common multiple?
Well let me ask, can $6n+5$ and $9n + 8$ have any factors in common?
Well, if $p|6n+5$ and $p|9n + 8$ then $p|(9n+8)-(6n+5)=3n+3$.  And in turn $p|(6n+5) - (3n+3) = 3n + 2$.  So $p|(3n+3) - (3n+2) = 1$.  So the only possible factor $(6n+5)$ and $(9n+8)$ can have in common is ... $1$.
Can I say with confidence that if two numbers have no factors in common, then their least common multiple is their product.
I can.  $\text{lcm}(a,b) = \frac {ab}{\gcd(a,b)}$ is a basic result.  The fact that these look like polynomials rather than constants doesn't actually make any difference.  We just have to worry that for different values of $n$ it might be possible that the two values may have different common divisors.  However that is not the case here.
A: Two numbers are co-prime when they share no common factor (ie when their Greatest Common Divisor is 1). The lowest common multiple of two co-prime numbers is their product. 
You need to show that the two numbers $9n+8$ and $6n+5$ are co-prime for any positive integer $n$, and then you're done immediately.
Hint: Consider some $k$ which divides both $9n+8$ and $6n+5$ and then deduce that $k=1$
A: You can do Euclidean algorithm to find the gcd.  This boils down to the fact that you can subtract multiples of one component of the gcd from the other repeatedly.
So:
$$(9n+8, 6n+5) = (3n+3,6n+5) = (3n+3, -1) = 1.$$ 
Since you know the gcd is $1$, you know the lcm is $(9n+8)(6n+5).$
A: You have to show that $(9n+8)$ and $(6n+5)$ are co-prime, i.e., their greatest common divisor is 1.
Assume that $k$ devides $6n+5$, i.e., $6n+5\bmod k=0$. Then 
\begin{align*}
9n+8&=3n+3\mod k\\
&=\frac{1}{2}\cdot2\cdot(3n+3)\mod k\\
&=\frac{1}{2}\cdot(6n+6)\mod k\\
&=\frac{1}{2}.
\end{align*}
If $k>1$ then $\frac{1}{2}$ is not zero and $k$ does not devide $9n+8$. Hence $(9n+8)$ and $(6n+5)$ are co-prime.
There probably is a more elegant proof but I do not see it right now.
Hope I could help :)
