extra tab

Oleksandr Kulkov · adamant-pwn · commit 19c7a40b422d · 2022-11-29T02:19:03.000+01:00
diff --git a/src/algebra/continued-fractions.md b/src/algebra/continued-fractions.md
@@ -678,22 +678,22 @@ One can further prove (and it was first done by Lagrange) that for arbitrary qua
         ```py
         # compute the continued fraction of sqrt(n)
         def sqrt(n):
-          n0 = math.floor(math.sqrt(n))
-          x, y, z = 1, 0, 1
-          a = []
-          def step(x, y, z):
-              a.append((x * n0 + y) // z)
-              t = y - a[-1]*z
-              x, y, z = -z*x, z*t, t**2 - n*x**2
-              g = math.gcd(x, math.gcd(y, z))
-              return x // g, y // g, z // g
-
-          used = dict()
-          for i in range(n):
-              used[x, y, z] = i
-              x, y, z = step(x, y, z)
-              if (x, y, z) in used:
-                  return a
+            n0 = math.floor(math.sqrt(n))
+            x, y, z = 1, 0, 1
+            a = []
+            def step(x, y, z):
+                a.append((x * n0 + y) // z)
+                t = y - a[-1]*z
+                x, y, z = -z*x, z*t, t**2 - n*x**2
+                g = math.gcd(x, math.gcd(y, z))
+                return x // g, y // g, z // g
+
+            used = dict()
+            for i in range(n):
+                used[x, y, z] = i
+                x, y, z = step(x, y, z)
+                if (x, y, z) in used:
+                    return a
         ```
 
     Using the same `step` function but different initial $x$, $y$ and $z$ it is possible to compute it for arbitrary $\frac{x+y \sqrt{n}}{z}$.
