/* From B = a mod |n|, A = |n| it follows that
*
* 0 <= B < A,
- * X*a == B (mod |n|),
+ * sign*X*a == B (mod |n|),
* -sign*Y*a == A (mod |n|).
*/
/*
* 0 < B < A,
- * (*) X*a == B (mod |n|),
+ * (*) sign*X*a == B (mod |n|),
* -sign*Y*a == A (mod |n|)
*/
* i.e.
* -sign*Y*a - D*A == B (mod |n|).
* Similarly, (*) translates into
- * X*a == A (mod |n|).
+ * sign*X*a == A (mod |n|).
*
* Thus,
- * -sign*Y*a - D*X*a == B (mod |n|),
+ * -sign*Y*a - D*sign*X*a == B (mod |n|),
* i.e.
- * -sign*(Y + D*X)*a == B (mod |n|).
+ * -sign*(Y + D*X)*a == B (mod |n|).
*
* So if we set (X, Y, sign) := (Y + D*X, X, -sign), we arrive back at
- * X*a == B (mod |n|),
+ * sign*X*a == B (mod |n|),
* -sign*Y*a == A (mod |n|).
* Note that X and Y stay non-negative all the time.
*/
}
/*
- * The while loop ends when
+ * The while loop (Euclid's algorithm) ends when
* A == gcd(a,n);
* we have
* -sign*Y*a == A (mod |n|),