证明任何自然数的平方要么是 3 的倍数，要么是 3 的倍数大一

数字系统是一种以某种方式使用符号和数字来表示数字的系统。人们可以将其视为一本数学语法书。一听到“数字”这个词 1,2,3,……。立即在脑海中浮现。数字系统定义了它们的值、要执行的操作和其他属性。每个号码都是独一无二的。它有很多变化。因此，它可以被认为是自然数、整数、偶数、奇数、素数、合数等。

1. 自然数- 它包含从 1 开始的数字。
2. 整数 - 它包含从 0 开始的数字。
3. 偶数——能被 2 整除的数。
4. 奇数——不能被 2 整除的数。
5. 素数——只能被 1 和自身整除的数字。即只有两个因素。
6. 合数– 能被 1 和自身和其他数整除的数。即两个以上的因素。

代数

代数是数学的一个分支，强调使用各种符号和可以对其进行的数学运算。在代数中，使用诸如字母表之类的符号，如 a、b、c 或 x，以及许多其他值未预定义的符号。它们的值不是固定的，因此称为变量。可以声明任意数量的变量并为它们定义特定的值。这些符号或变量可以在算术运算（如加法、减法、乘法和除法）的帮助下进行操作。让我们处理代数的一种应用，即代数表达式。

代数表达式由系数、项、变量、常数等组成，人们必须通过数学运算来确定未知变量的值。

自然数

自然数是从 1 到无穷大的计数数，它们用于计数目的。例如 1,2,3,4,…。 , 是自然数。基本上，自然数是实数系统的一个组成部分。

However, Natural numbers doesn’t include zero, fraction, decimal, negative numbers.

一组自然数用 N 表示，使得 N = {1, 2, 3, 4, 5, 6, 7, 8,….}。

用代数证明任何自然数的平方要么是 3 的倍数，要么是 3 的倍数大一。

解决方案：

Suppose, we have a natural number  N.

Consider three cases where N is in form of  3*N, 3*N+1, 3*N+2.

Squaring, we get

• The square of 1st term is 9N²  which is a multiple of 3.
• Square of 2nd term is:

9N² + 6N +1 = 3(3N² + 2N) + 1 which is one more than a multiple of 3.

• Square of 3rd term is:

9N² + 12N + 4 =  9N² +12N + 3 + 1 = 3(3N² + 4N + 1) +1 which is one more than multiple of 3.

Hence, it can be said that for N²  to be multiple of 3, N should be in form of 3N, 3N+1, 3N+2, and so on.

Example:

Square Number	Remainder when divided by 3
22 = 4 = 3 × 1 + 1	1
32 = 9 = 3 × 3 + 0	0
42 = 16 = 3 × 5 + 1	1
52 = 25 = 3 × 8 + 1	1
62 = 36 = 3 × 12 + 0	0
72 = 49 = 3 × 16 + 1	1

Hence, the square of any natural number is either a multiple of 3 or one more than a multiple of 3.

类似问题

问题1：证明两个连续平方数之和是奇数？

解决方案：

Let us assume that N and N + 1 are two consecutive numbers where N is any integer.

We have to determine their sum. So,

Sum = N² + (N + 1)² = N² + N² + 2N + 1 = 2N² + 2N + 1 = 2(N² + N) +1

The definition of odd number is one more than multiple of 2.

 The sum is one more than multiple of 2. So, their sum is odd.

Example:

Input N = 2 and N + 1 = 3

Output 2² + 3² = 4+ 9 = 13

Conclusion The sum is odd.

问题 2：证明如果 m 不是自然数的平方，则 √m 是无理数。

解决方案：

Let m be any positive integer such that there is no m = x² where x is an integer.

We assume that √m is a rational number. Then it can be written as:

√m = p/q  ⇢ (1)

Where p and q have no common factor other than 1.

Squaring both sides of Equation (1)

m = p² / q².

Since n is a positive integer and p and q have no common factor besides (1). Let q = 1.

m = p² 

This result is contradictory to our assumption that m is not a square of a number. Hence, √m is irrational.

问题 3：证明如果 n 是自然数，则 √n 是无理数或自然数。

解决方案：

Let n be any natural number.

Case 1: n is a perfect square (1, 4, 9, 16, etc.)

Natural number (n)	The root of a number (√n)	Definition of √n
1	1	Well – defined
4	2	Well – defined
9	3	Well – defined
36	6	Well – defined

Hence, √n is a natural number.

Case 2: n is not a perfect square (√2, √3, √7, etc.)

Natural number(n)	The root of a number(√n)	Definition of √n
2	√2	Not Well – defined
3	√3	Not Well – defined
7	√3	Not Well – defined
19	√19	Not Well – defined

Hence, √n is an irrational number.

Conclusion √n is either a natural number or an irrational number.

问题4：证明两个完全平方数的乘积也是一个完全平方。

解决方案：

Let p and q be any two positive integers.

Since both p and q are perfect squares, we can write

p = a²

q = b²

Where a and b are any two integers.

p × q = a² × b² = (a × a) × (b × b) = (a × b) × (a × b) = (ab) × (ab) = (ab)²

Hence, The result is a perfect square of ab.

Conclusion The product of two perfect square numbers is also perfect square.

Example:

Input p=4 and q=9

Output pq= 4*9 = 36 =6²

Which is a perfect square.