pythagorean-triplet: Faster tests with a fast example (#262)

This shaves 25% off the test time for me.

exercises/pythagorean-triplet/example.jl

@@ -1,13 +1,27 @@
-function pythagorean_triplets(input)
-    triplets = []
-    for a in 1:(ceil(input/2))
-        for b in a:(ceil((input-a)/2))
-            c = input - a - b
-            triplet = sort([a, b, c])
-            if (triplet[1]^2 + triplet[2]^2 == triplet[3]^2)
-                push!(triplets, Tuple(Int.(triplet)))
-            end
+"""
+    pythagorean_triplets(n)
+
+Find all positive integer triplets `(a, b, c)` s.t. `a + b + c = n` and `a < b < c` and `a^2 + b^2 == c^2`.
+"""
+# cmcaine's answer, with thanks to akshu3398.
+function pythagorean_triplets(n)
+    triplets = NTuple{3, Int}[]
+    # Lower bound because the smallest triple is 3, 4, 5.
+    # Upper bound implied by a < b < c && a + b + c == n.
+    for a in 3:cld(n, 3) - 1
+        # Derived by eliminating c from these simultaneous
+        # equations and solving for b:
+        #     a^2 + b^2 = c^2
+        #     a + b + c = n
+        b = (n^2 - 2n * a) / (2 * (n - a))
+        if a < b && isinteger(b)
+            c = n - a - b
+            # Proof that b < c:
+            # Let b = c. Then a^2 + b^2 = c^2 ≡ a^2 = 0 but we know that a ≠ 0.
+            # Let b > c. a^2 + b^2 = c^2 ≡ a^2 = c^2 - b^2.
+            #    If b > c, a^2 < 0, but that is impossible for real numbers.
+            push!(triplets, (a, b, c))
         end
     end
-    return(unique(triplets))
+    return sort!(triplets)
 end