$nx - my = 0$, find smallest positive integer values of $nx$ and $my$ where this is true My first question on math stackexchange.
I have two numbers, $n$ and $m$.  Their values will vary but I always know what they are and they're always positive integers (and multiples of 2, but that doesn't make much difference).
I want to find the smallest number both $n$ and $m$ will perfectly divide into.
E.g.$$\begin{align*}n&=32\\m&=144\end{align*}$$
Smallest number they will both divide into is $288$.
I can see how to work this out iteratively ($n*m$, then keep dividing by 2 until you find the first number they both won't divide into, the previous value is the desired value) but I can't see how to calculate it directly (and I need to).
For those who ae wondering what this is for, it's do with memory allocation on an ARM platform.  I want to be cache line aligned and also exclusive reservation granuale aligned.  There, that helped ;-)
So if the cache line width is 32 bytes and the ERG is 144 bytes, I want to allocate with alignment on 288 bytes.
What it amounts to is this;
$$nx - my = 0$$
Except that defines many values and I want the smallest value of $nx$ and $my$ where this is true.
$$\begin{align*}
    nx &= my\\
    \frac{nx}{ my} &= 1\\
    \left(\frac{n}{m}\right)\left(\frac{x}{y}\right) &= 1
\end{align*}$$
I know $n$ and $m$, so $\frac{n}{m} = c$
$$\frac{x}{y}=c$$
And I see that somewhere I need to find more information because what I have can't solve the question.  I can't see how to quantify my wish for the smallest values into the equation.
 A: The problem you are describing is called least common multiple. I don't think there exists closed form of LCM, since it would give us closed solution of greatest common divisor. 
A: First, find the greatest common divisor of $m$ and $n$. This can be done using the Euclidean algorithm, which is fast and computationally efficient.
Once you know the greatest common divisor $d$ of $m$ and $n$, we can write $m=dm'$, $n=dn'$, and the least common multiple of $m$ and $n$ is $dm'n'$. So $x=m'$, $y=n'$ and we have $xn-ym=0$. This is the smallest values that works.
A: Call $(p_i)_{i\geqslant1}$ the prime integers and consider two integers $n$ and $m$ whose  decompositions into products of prime factors are $n=\prod\limits_{i\geqslant1}p_i^{c_i(n)}$ and $m=\prod\limits_{i\geqslant1}p_i^{c_i(m)}$. 
Then the greatest common divisor of $n$ and $m$ is $\gcd(n,m)=\prod\limits_{i\geqslant1}p_i^{\min\{c_i(n),c_i(m)\}}$ and the lowest common multiple of $n$ and $m$ (which coincides with your smallest $nx=my$) is $\text{lcm}(n,m)=\prod\limits_{i\geqslant1}p_i^{\max\{c_i(n),c_i(m)\}}$.
A: Hint $\rm\ n\,x = m\ y\iff x/y = m/n\ $ so least $\rm\, x,y\, $ arise from $\rm\,m/n\,$ in least terms $\rm = (m/d)/(n/d)\,$ where $\rm\,d = gcd(m,n).\, $ So the least solution is $\rm\, n\,x = n m/d,\,$ i.e. $\rm\,lcm(n,m) = nm/gcd(n,m).$
