Show that the square of any positive odd integer is of the form 4m+1 for some integer n.

Hint: Here, to prove that the square of any positive odd number is of the form 4m+1, we are going to use the Euclid’s Division Lemma. This theorem states that if we have two positive integers a and b then there exists unique integers q and r that satisfy the condition $ a = bq + r $ where $ 0 \leqslant r < b $ . Using this theorem, we will be proving the given statement.

Complete step-by-step answer:
In this question, we have to prove that the square of any positive odd number is of the form 4m+1.
For proving this, we are going to use the Euclid’s Division Lemma. So, first of all let us see what this theorem is.
According to this theorem, if we have two positive integers a and b, then there exists unique integers q and r that satisfy the condition $ a = bq + r $ where $ 0 \leqslant r < b $ . This is known as the Euclid’s Division Lemma.
So, applying Euclid’s Division Lemma with a, b, q, r and $ b = 4 $ , we get equation
 $ \Rightarrow a = bq + r $ , $ 0 \leqslant r < b $
 $ \Rightarrow a = 4q + r $ , $ 0 \leqslant r < 4 $
So, we need to take 4 values of r.

Case 1: When $ r = 0 $
 $ \Rightarrow a = 4q $
Therefore, its square will be
 $ \Rightarrow {a^2} = 16{q^2} = 4\left( {4{q^2}} \right) = 4m $
Here, we have taken $ 4{q^2} = m $

Case 2: When $ r = 1 $
 $ \Rightarrow a = 4q + 1 $
Therefore, its square will be
 $ \Rightarrow {a^2} = {\left( {4q + 1} \right)^2} = 16{q^2} + 8q + 1 = 4\left( {4{q^2} + 2q} \right) + 1 = 4m + 1 $
Here, we have taken $ \left( {4{q^2} + 2q} \right) = m $

Case 3: When $ r = 2 $
 $ \Rightarrow a = 4q + 2 $
Therefore, its square will be
 $ \Rightarrow {a^2} = {\left( {4q + 2} \right)^2} = 16{q^2} + 16q + 4 = 4\left( {4{q^2} + 4q + 1} \right) = 4m $
Here, we have taken $ \left( {4{q^2} + 4q + 1} \right) = m $

Case 4: When $ r = 3 $
 $ \Rightarrow a = 4q + 3 $
Therefore, its square will be
 $ \Rightarrow {a^2} = {\left( {4q + 3} \right)^2} = 16{q^2} + 24q + 9 = 4\left( {4{q^2} + 12q + 2} \right) + 1 = 4m + 1 $
Here, we have taken $ \left( {4{q^2} + 12q + 2} \right) = m $ .
Hence, we can see that when the value of r is odd, we get the form 4m+1 and when the value of r is even, we get the form 4m.

Note: Here the most important part of the question is taking the value of b equal to 4. As we have to prove the form 4m+1, we need to take b as 4 only. Another important form is taking out 4 common in the equations obtained by squaring. No matter what can be taken out of common, we need to take out 4 as common only and then substitute the remaining part with any variable.