Understanding Montgomery's parameterization of elliptic curves I'm having a bit of trouble understanding the translation of affine  coordinates to projective coordinates in Montgomery curve ECM. Would be very thankful if someone could explain it by expanding the derivation. 


  From: Speeding the Pollard and Elliptic Curve Methods
  of Factorization page: 261.
  $$\begin{equation}\tag{10.3.1.1}By^2 = x^3  + Ax^2 + x\end{equation}$$
  
  Let $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ be two points in the
  curve, with $x_1 \neq x_2$ and $x_1x_2 \neq 0$.  Then $P_1 + P_2 = (x_3, y_3)$ satisfies
  
  $$x_3 = B[(y_1 -  y_2)/(x_1 - x_2)]^2 - A - x_1 - x_2,$$
   $$\begin{align}x_3(x_1 - x_2)^2 &amp; = B(y_1 -  y_2)^2 - (A + x_1 +
 x_2)(x_1 - x_2)^2 \\ &amp; = -2By_1y_2 + x_1x_2(x_1 + x_2 + 2A) + x_1 +
 x_2 \\ &amp; = B(x_2y_1 - x_1y_2)^2/x_1x_2. \end{align}$$
  
  Similarly, $P_1 - P_2 = (x_4, y_4)$ satisfies $$x_4(x_1 - x_2)^2 =
 B(x_2y_1 + x_1y_2)^2/x_1x_2.$$ Multiply these equations and use
   (10.3.1.1) to obtain $$x_3x_4(x_1 - x_2)^2 = (x_1x_2 - 1)^2$$ after
  division by $(x_1 - x_2)^2$. This equation remains valid if $x_1x_2 = 0$. If $P_1 = P2$, a similar derivation yields
  
  $$4x_1x_3(x_1^2 + Ax_1 + 1) = (x_1^2 - 1)^2$$  These equations
  reference only the $x_i$, not the $y_i$. Fortunately ECM does not
  require us to compute the $y_i$.


(I  don't understand everything below this point. Could someone help me expand these derivations? Seems like it is oversimplified)


  Let $P$ be an arbitrary point on the curve and let the x-coordinate of
  $nP$ be the rational number $X_n/Z_n$. From the ratio
  $(X_{m-n}:Z_{m-n}),(X_m:Z_m) and (X_n:Z_n)$, one  can compute the
  ratio $(X_{m-n}:Z_{m-n})$ via the addition formula.
  
  $$X_{m+n} \leftarrow Z_{m-n}(X_mX_n - Z_mZ_n)^2\\ Z_{m+n} \leftarrow X_{m-n}(X_mZ_n - Z_mX_n)^2$$
  
  if $mP \neq nP$, and  via the duplication  formula $$\begin{align}
X_{2n} &amp; \leftarrow (X_n^2 - Z_n^2)^2, \\ 
Z_{2n} &amp; \leftarrow 4X_nZ_n(X_n^2 + AX_nZ_n + Z_n^2)
\end{align}$$
  
  The costs drop if we store the ratios $(X_m:Z_m:X_m+Z_m:X_m-Z_m)$ and rewrite the formulae as (the right sides of the addition formula have 
  been multiplied by 4)
  
  $$\begin{align}
X_{m+n} &amp; \leftarrow Z_{m-n}[(X_m - Z_m)(X_n + Z_n) + (X_m + Z_m)(X_n - Z_n)]^2 \\ 
Z_{m+n} &amp; \leftarrow X_{m-n}[(X_m - Z_m)(X_n + Z_n) - (X_m + Z_m)(X_n - Z_n)]^2 \\
\end{align}$$
  and
  
  $$\begin{align}
4X_nZ_n &amp; = (X_n + Z_n)^2 - (X_n - Z_n) \\
X_{2n} &amp;  \leftarrow (X_n + Z_n)^2(X_n - Z_n)^2 \\
Z_{2n} &amp; \leftarrow (4X_nZ_n)((X_n - Z_n)^2 + ((A + 2)/4)(4X_nZ_n))
\end{align}$$
Write $(x_i, y_i) = (X_i : Y_i : Z_i)$, so that $x_i = X_i/Z_i$ and $y_i = X_i/Z_i$, where $Z_i \ne 0$ is arbitrary.  (If you are not familiar with projective coordinates or you like visuals, see an illustration of projective coordinates on a real elliptic curve.)

Let's take the doubling formula for example, where $(x_3, y_3) = (x_1, y_1) + (x_1, y_1) = [2](x_1, y_1)$:

\begin{equation*}
  4 x_1 x_3 ({x_1}^2 + A x_1 + 1) = ({x_1}^2 - 1)^2 \tag{p. 261, second display}
\end{equation*}

so that

\begin{equation*}
  x_3 = \frac{({x_1}^2 - 1)^2}{4 x_1 ({x_1}^2 + A x_1 + 1)},
\end{equation*}

which in projective coordinates is

\begin{align*}
  \frac{X_3}{Z_3}
   &amp;= \frac{\Bigl(\bigl(\frac{X_1}{Z_1}\bigr)^2 - 1\Bigr)^2}
           {4 \frac{X_1}{Z_1}
              \Bigl(\bigl(\frac{X_1}{Z_1}\bigr)^2 + A \frac{X_1}{Z_1} + 1\Bigr)} \\
   &amp;= \frac{{Z_1}^4}{{Z_1}^4}
      \cdot
      \frac{\Bigl(\bigl(\frac{X_1}{Z_1}\bigr)^2 - 1\Bigr)^2}
           {4 \frac{X_1}{Z_1}
              \Bigl(\bigl(\frac{X_1}{Z_1}\bigr)^2 + A \frac{X_1}{Z_1} + 1\Bigr)} \\
   &amp;= \frac{\bigl({Z_1}^2\bigr)^2 \Bigl(\bigl(\frac{X_1}{Z_1}\bigr)^2 - 1\Bigr)^2}
           {4 X_1 Z_1 {Z_1}^2
            \Bigl(\bigl(\frac{X_1}{Z_1}\bigr)^2 + A \frac{X_1}{Z_1} + 1\Bigr)} \\
   &amp;= \frac{\bigl({X_1}^2 - {Z_1}^2\bigr)^2}
           {4 X_1 Z_1 \bigl({X_1}^2 + A X_1 Z_1 + {Z_1}^2\bigr)}.
\end{align*}

From this, we can read off the numerator and denominator:

\begin{equation*}
  X_3 = \bigl({X_1}^2 - {Z_1}^2\bigr)^2
  \qquad\text{and}\qquad
  Z_3 = 4 X_1 Z_1 \bigl({X_1}^2 + A X_1 Z_1 + {Z_1}^2\bigr).
\end{equation*}

Obviously, we can multiply both of them by the same arbitrary nonzero factor too, but there's no need here.

For the addition formula that is not doubling, you're trying to write $[m + n]P = [m]P + [n]P$ in terms of $X$ and $Z$ coordinates for $[m]P$, $[n]P$, and $[m - n]P = [m]P - [n]P$.  That is, you have $x_1 = x([m]P)$, $x_2 = x([n]P)$, and $x_4 = x([m - n]P)$, and you're trying to find $x_3 = x([m + n]P)$.  Use the top equation on p. 261, $$x_3 x_4 (x_1 - x_2)^2 = (x_1 x_2 - 1)^2,$$ and again write it out in terms of $x_i = X_i/Z_i$.

The last part is a matter of tidying to reduce the number of distinct intermediate quantities, e.g. using the observation that

\begin{multline*}
  {X_1}^2 + A X_1 Z_1 + {Z_1}^2
   = {X_1}^2 - 2 X_1 Z_1 + {Z_1}^2 + 2 X_1 Z_1 + (A/4) 4 X_1 Z_1 \\
   = (X_1 - Z_1)^2 + \frac{A + 2}{4} 4 X_1 Z_1
\end{multline*}

to write the whole doubling formula in terms of $X_1 \pm Z_1$ for a total cost of 4A + 2S + 3M.