✨ feat(euler_45)：添加欧拉项目第45题解决方案

📝 docs(euler_45)：添加详细数学推导和佩尔方程说明文档

solutions/0045.TrianPentaHexa/euler_45.py

@@ -0,0 +1,86 @@
+"""
+Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
+    Triangle: T_n = n(n+1)/2
+    Pentagonal: P_n = n(3n-1)/2
+    Hexagonal: H_n = n(2n-1)
+
+It can be verified that T_285 = P_165 = H_143 = 40755.
+
+Find the next triangle number that is also pentagonal and hexagonal.
+"""
+
+import time
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        print(f"{func.__name__} taken: {end_time - start_time:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+def is_pentagonal(x):
+    """检查一个数是否为五边形数"""
+    # 解方程 n(3n-1)/2 = x => 3n² - n - 2x = 0
+    # 判别式 Δ = 1 + 24x 必须是完全平方数
+    delta = 1 + 24 * x
+    sqrt_delta = int(delta**0.5)
+    if sqrt_delta * sqrt_delta != delta:
+        return False
+    # 并且 (1 + sqrt(Δ)) ≡ 0 mod 6
+    return (1 + sqrt_delta) % 6 == 0
+
+
+def is_hexagonal(x):
+    """检查一个数是否为六边形数"""
+    # 解方程 n(2n-1) = x => 2n² - n - x = 0
+    # 判别式 Δ = 1 + 8x 必须是完全平方数
+    delta = 1 + 8 * x
+    sqrt_delta = int(delta**0.5)
+    if sqrt_delta * sqrt_delta != delta:
+        return False
+    # 并且 (1 + sqrt(Δ)) ≡ 0 mod 4
+    return (1 + sqrt_delta) % 4 == 0
+
+
+@timer
+def main() -> None:
+    n = 286
+    while True:
+        triangle = n * (n + 1) // 2
+
+        # 直接使用已有的验证函数，避免重复计算
+        if is_pentagonal(triangle) and is_hexagonal(triangle):
+            # 找到后计算对应的索引
+            pent_index = (1 + int((1 + 24 * triangle) ** 0.5)) // 6
+            hex_index = (1 + int((1 + 8 * triangle) ** 0.5)) // 4
+            print(f"Answer: T({n})=P({pent_index})=H({hex_index})={triangle}")
+            break
+
+        n += 1
+
+
+@timer
+def main_quick_math(nth: int = 2) -> None:
+    u = 5
+    v = 3
+    sn = 0
+    while True:
+        u, v = (2 * u + 3 * v), (u + 2 * v)
+        if u % 6 == 5 and v % 4 == 3:
+            k = (u + 1) // 6
+            j = (v + 1) // 4
+            n = 2 * j - 1
+            sn += 1
+            if sn == nth:
+                print(f"Answer: T({n})=P({k})=H({j})={n * (n + 1) // 2}")
+                break
+
+
+if __name__ == "__main__":
+    main()
+    main_quick_math()