Question

Prove that the square of any odd $ {Z^ + } $ is of the form $ 4q + 1 $ for some integer $ q $

Answer 1

Hint: First write the odd positive integer with the help of Euclidean algorithm by taking the dividend as $ 2 $ . Then take the square of that type of number and represent it as it is asked to be prove.

Complete step-by-step answer:
Any odd positive integer when divided by $ 2 $ gives the remainder as $ 1 $ .
Let $ a $ be any odd positive integer.
So with the help of the Euclidean algorithm the number $ a $ can be represented as $ a = 2r + 1 $ , where $ r $ is any positive integer.
Now take the square of the number $ a $ .
  $ {a^2} = {\left( {2r + 1} \right)^2} $
Use the property $ {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab $ to simplify the above expression.
 $
  {a^2} = {\left( {2r + 1} \right)^2} \\
   = {\left( {2r} \right)^2} + {\left( 1 \right)^2} + 2 \times 2r \times 1 \\
   = 4{r^2} + 1 + 4r \;
  $
Now rearrange the terms of the above expression and make the number of types as required in the question.
 $
  {a^2} = 4{r^2} + 1 + 4r \\
   = 4\left( {{r^2} + r} \right) + 1 \\
   = 4q + 1 \;
  $
Where, the term $ {r^2} + r $ is taken to be equal to $ q $ .
 $ r $ is a positive integer. So, $ {r^2} + r $ is also a positive integer.
So, any odd positive integer can be represented in the form of $ 4q + 1 $ where $ q $ is some integer.
Hence, the required result is proved.

Note: Take the combination of only those terms which are required in the proof. Take others on the side and make the result required. Any odd positive integer when divided by $ 2 $ always leaves $ 1 $ as remainder. Euclidean algorithm states that is a number $ a $ when divided by a number $ b $ leaves the remainder $ r $ and quotient as $ q $ , then the number $ a $ can be represented as $ a = bq + r $ .