Show that the square of any positive integer is either of the form 4q or 4q + 1 form some integer q

Answer

To show that the square of any positive integer is either of the form \(4q\) or \(4q + 1\) for some integer \(q\), we can use the concept of division and the properties of integers. Here’s a step-by-step solution:

### Step 1: Understanding the Positive Integer
Let \(A\) be any positive integer. According to the division lemma, any integer can be expressed in the form:
\[ A = 4M + R \]
where \(M\) is some integer and \(R\) is the remainder when \(A\) is divided by \(4\). The possible values for \(R\) (the remainder) can be \(0\), \(1\), \(2\), or \(3\).

### Step 2: Case Analysis Based on Remainder
We will analyze the cases based on the possible values of \(R\).

#### Case 1: \(R = 0\)
If \(R = 0\), then:
\[ A = 4M \]
Now, squaring \(A\):
\[ A^2 = (4M)^2 = 16M^2 = 4(4M^2) \]
This shows that \(A^2\) is of the form \(4q\) where \(q = 4M^2\).

#### Case 2: \(R = 1\)
If \(R = 1\), then:
\[ A = 4M + 1 \]
Now, squaring \(A\):
\[ A^2 = (4M + 1)^2 = 16M^2 + 8M + 1 = 4(4M^2 + 2M) + 1 \]
This shows that \(A^2\) is of the form \(4q + 1\) where \(q = 4M^2 + 2M\).

#### Case 3: \(R = 2\)
If \(R = 2\), then:
\[ A = 4M + 2 \]
Now, squaring \(A\):
\[ A^2 = (4M + 2)^2 = 16M^2 + 16M + 4 = 4(4M^2 + 4M + 1) \]
This shows that \(A^2\) is of the form \(4q\) where \(q = 4M^2 + 4M + 1\).

#### Case 4: \(R = 3\)
If \(R = 3\), then:
\[ A = 4M + 3 \]
Now, squaring \(A\):
\[ A^2 = (4M + 3)^2 = 16M^2 + 24M + 9 = 4(4M^2 + 6M + 2) + 1 \]
This shows that \(A^2\) is of the form \(4q + 1\) where \(q = 4M^2 + 6M + 2\).

### Conclusion
From the analysis of all cases, we can conclude that the square of any positive integer \(A\) is either of the form \(4q\) or \(4q + 1\) for some integer \(q\).