feat: add solutions to lc problem: No.0633

solution/0600-0699/0633.Sum of Square Numbers/README.md

@@ -193,4 +193,138 @@ impl Solution {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### 方法二：数学
+
+这个问题实际上是关于一个数能否表示为两个平方数之和的条件。这个定理可以追溯到费马（Fermat）和欧拉（Euler），它在数论中是一个经典结果。
+
+具体来说，这个定理可以表述为：
+
+**一个正整数 \( n \) 能表示为两个平方数之和的充要条件是：\( n \) 的所有形如 \( 4k + 3 \) 的素数因子的幂次均为偶数。**
+
+这意味着，如果我们将 $n$ 分解成素数因子乘积的形式，即 $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$，其中 $p_i$ 是素数且 $e_i$ 是它们对应的幂次，那么 $n$ 可以表示为两个平方数之和，当且仅当所有 $4k + 3$ 形式的素数因子 $p_i$ 的幂次 $e_i$ 都是偶数。
+
+更正式地，假设 $p_i$ 是形如 $4k + 3$ 的素数，则对于每一个这样的 $p_i$，要求 $e_i$ 是偶数。
+
+例如：
+
+-   数字 $13$ 是素数，且 $13 \equiv 1 \pmod{4}$，因此它可以表示为两个平方数之和，即 $13 = 2^2 + 3^2$。
+-   数字 $21$ 分解为 $3 \times 7$，其中 $3$ 和 $7$ 都是形如 $4k + 3$ 的素数因子，并且它们的幂次都是 $1$（奇数），因此 $21$ 不能表示为两个平方数之和。
+
+总结起来，这个定理在数论中非常重要，用于判断一个数是否可以表示为两个平方数之和。
+
+时间复杂度 $O(\sqrt{c})$，其中 $c$ 是给定的非负整数。空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def judgeSquareSum(self, c: int) -> bool:
+        for i in range(2, int(sqrt(c)) + 1):
+            if c % i == 0:
+                exp = 0
+                while c % i == 0:
+                    c //= i
+                    exp += 1
+                if i % 4 == 3 and exp % 2 != 0:
+                    return False
+        return c % 4 != 3
+```
+
+#### Java
+
+```java
+class Solution {
+    public boolean judgeSquareSum(int c) {
+        int n = (int) Math.sqrt(c);
+        for (int i = 2; i <= n; ++i) {
+            if (c % i == 0) {
+                int exp = 0;
+                while (c % i == 0) {
+                    c /= i;
+                    ++exp;
+                }
+                if (i % 4 == 3 && exp % 2 != 0) {
+                    return false;
+                }
+            }
+        }
+        return c % 4 != 3;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    bool judgeSquareSum(int c) {
+        int n = sqrt(c);
+        for (int i = 2; i <= n; ++i) {
+            if (c % i == 0) {
+                int exp = 0;
+                while (c % i == 0) {
+                    c /= i;
+                    ++exp;
+                }
+                if (i % 4 == 3 && exp % 2 != 0) {
+                    return false;
+                }
+            }
+        }
+        return c % 4 != 3;
+    }
+};
+```
+
+#### Go
+
+```go
+func judgeSquareSum(c int) bool {
+	n := int(math.Sqrt(float64(c)))
+	for i := 2; i <= n; i++ {
+		if c%i == 0 {
+			exp := 0
+			for c%i == 0 {
+				c /= i
+				exp++
+			}
+			if i%4 == 3 && exp%2 != 0 {
+				return false
+			}
+		}
+	}
+	return c%4 != 3
+}
+```
+
+#### TypeScript
+
+```ts
+function judgeSquareSum(c: number): boolean {
+    const n = Math.floor(Math.sqrt(c));
+    for (let i = 2; i <= n; ++i) {
+        if (c % i === 0) {
+            let exp = 0;
+            while (c % i === 0) {
+                c /= i;
+                ++exp;
+            }
+            if (i % 4 === 3 && exp % 2 !== 0) {
+                return false;
+            }
+        }
+    }
+    return c % 4 !== 3;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->

solution/0600-0699/0633.Sum of Square Numbers/README_EN.md

@@ -191,4 +191,138 @@ impl Solution {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### Solution 2: Mathematics
+
+This problem is essentially about the conditions under which a number can be expressed as the sum of two squares. This theorem dates back to Fermat and Euler and is a classic result in number theory.
+
+Specifically, the theorem can be stated as follows:
+
+> A positive integer $n$ can be expressed as the sum of two squares if and only if all prime factors of $n$ of the form $4k + 3$ have even powers.
+
+This means that if we decompose $n$ into the product of its prime factors, $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, where $p_i$ are primes and $e_i$ are their corresponding exponents, then $n$ can be expressed as the sum of two squares if and only if all prime factors $p_i$ of the form $4k + 3$ have even exponents $e_i$.
+
+More formally, if $p_i$ is a prime of the form $4k + 3$, then for each such $p_i$, $e_i$ must be even.
+
+For example:
+
+-   The number $13$ is a prime and $13 \equiv 1 \pmod{4}$, so it can be expressed as the sum of two squares, i.e., $13 = 2^2 + 3^2$.
+-   The number $21$ can be decomposed into $3 \times 7$, where both $3$ and $7$ are prime factors of the form $4k + 3$ and their exponents are $1$ (odd), so $21$ cannot be expressed as the sum of two squares.
+
+In summary, this theorem is very important in number theory for determining whether a number can be expressed as the sum of two squares.
+
+The time complexity is $O(\sqrt{c})$, where $c$ is the given non-negative integer. The space complexity is $O(1)$.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def judgeSquareSum(self, c: int) -> bool:
+        for i in range(2, int(sqrt(c)) + 1):
+            if c % i == 0:
+                exp = 0
+                while c % i == 0:
+                    c //= i
+                    exp += 1
+                if i % 4 == 3 and exp % 2 != 0:
+                    return False
+        return c % 4 != 3
+```
+
+#### Java
+
+```java
+class Solution {
+    public boolean judgeSquareSum(int c) {
+        int n = (int) Math.sqrt(c);
+        for (int i = 2; i <= n; ++i) {
+            if (c % i == 0) {
+                int exp = 0;
+                while (c % i == 0) {
+                    c /= i;
+                    ++exp;
+                }
+                if (i % 4 == 3 && exp % 2 != 0) {
+                    return false;
+                }
+            }
+        }
+        return c % 4 != 3;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    bool judgeSquareSum(int c) {
+        int n = sqrt(c);
+        for (int i = 2; i <= n; ++i) {
+            if (c % i == 0) {
+                int exp = 0;
+                while (c % i == 0) {
+                    c /= i;
+                    ++exp;
+                }
+                if (i % 4 == 3 && exp % 2 != 0) {
+                    return false;
+                }
+            }
+        }
+        return c % 4 != 3;
+    }
+};
+```
+
+#### Go
+
+```go
+func judgeSquareSum(c int) bool {
+	n := int(math.Sqrt(float64(c)))
+	for i := 2; i <= n; i++ {
+		if c%i == 0 {
+			exp := 0
+			for c%i == 0 {
+				c /= i
+				exp++
+			}
+			if i%4 == 3 && exp%2 != 0 {
+				return false
+			}
+		}
+	}
+	return c%4 != 3
+}
+```
+
+#### TypeScript
+
+```ts
+function judgeSquareSum(c: number): boolean {
+    const n = Math.floor(Math.sqrt(c));
+    for (let i = 2; i <= n; ++i) {
+        if (c % i === 0) {
+            let exp = 0;
+            while (c % i === 0) {
+                c /= i;
+                ++exp;
+            }
+            if (i % 4 === 3 && exp % 2 !== 0) {
+                return false;
+            }
+        }
+    }
+    return c % 4 !== 3;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->

solution/0600-0699/0633.Sum of Square Numbers/Solution2.cpp

@@ -0,0 +1,19 @@
+class Solution {
+public:
+    bool judgeSquareSum(int c) {
+        int n = sqrt(c);
+        for (int i = 2; i <= n; ++i) {
+            if (c % i == 0) {
+                int exp = 0;
+                while (c % i == 0) {
+                    c /= i;
+                    ++exp;
+                }
+                if (i % 4 == 3 && exp % 2 != 0) {
+                    return false;
+                }
+            }
+        }
+        return c % 4 != 3;
+    }
+};

solution/0600-0699/0633.Sum of Square Numbers/Solution2.go

@@ -0,0 +1,16 @@
+func judgeSquareSum(c int) bool {
+	n := int(math.Sqrt(float64(c)))
+	for i := 2; i <= n; i++ {
+		if c%i == 0 {
+			exp := 0
+			for c%i == 0 {
+				c /= i
+				exp++
+			}
+			if i%4 == 3 && exp%2 != 0 {
+				return false
+			}
+		}
+	}
+	return c%4 != 3
+}

solution/0600-0699/0633.Sum of Square Numbers/Solution2.java

@@ -0,0 +1,18 @@
+class Solution {
+    public boolean judgeSquareSum(int c) {
+        int n = (int) Math.sqrt(c);
+        for (int i = 2; i <= n; ++i) {
+            if (c % i == 0) {
+                int exp = 0;
+                while (c % i == 0) {
+                    c /= i;
+                    ++exp;
+                }
+                if (i % 4 == 3 && exp % 2 != 0) {
+                    return false;
+                }
+            }
+        }
+        return c % 4 != 3;
+    }
+}

solution/0600-0699/0633.Sum of Square Numbers/Solution2.py

@@ -0,0 +1,11 @@
+class Solution:
+    def judgeSquareSum(self, c: int) -> bool:
+        for i in range(2, int(sqrt(c)) + 1):
+            if c % i == 0:
+                exp = 0
+                while c % i == 0:
+                    c //= i
+                    exp += 1
+                if i % 4 == 3 and exp % 2 != 0:
+                    return False
+        return c % 4 != 3

solution/0600-0699/0633.Sum of Square Numbers/Solution2.ts

@@ -0,0 +1,16 @@
+function judgeSquareSum(c: number): boolean {
+    const n = Math.floor(Math.sqrt(c));
+    for (let i = 2; i <= n; ++i) {
+        if (c % i === 0) {
+            let exp = 0;
+            while (c % i === 0) {
+                c /= i;
+                ++exp;
+            }
+            if (i % 4 === 3 && exp % 2 !== 0) {
+                return false;
+            }
+        }
+    }
+    return c % 4 !== 3;
+}