Show that the square of an odd positive integer is of the form $8q+1$, for some integer q.

Given: Statement "Square of an odd positive integer is of the form $8q\ +\ 1$, for some integer q".

To prove: Here we have to prove the given statement.

Solution:

Let 'a' be any positive integer.

So, by Euclid's division lemma:

$a\ =\ bq\ +\ r$, where $0\ \underline{< }\ r\ <\ b$.

Here, $b\ =\ 8$. Then,

$a\ =\ 8q\ +\ r$, where $0\ \underline{< }\ r\ <\ 8$.

But according to the question we have to take a square of an odd positive integer, then  $r\ =\ 1,\ 3,\ 5,\ 7$.

Taking, $r\ =\ 1$:

$a\ =\ bq\ +\ 1$

Squaring both sides:

$a^2\ =\ (8q\ +\ 1)^2$ 

$a^2\ =\ 64q^2\ +\ 16q\ +\ 1$

$a^2\ =\ 8(8q^2\ +\ 2q)\ +\ 1$

$a^2\ =\ 8m\ +\ 1$  

Where, $m\ =\ 8q^2\ +\ 2q$.

Taking, $r\ =\ 3$,

$a\ =\ bq\ +\ 3$ 

Squaring both sides:

$a^2=(8q+3)^2$ 

$a^2\ =\ 64q^2\ +\ 48q\ +\ 9$

$a^2\ =\ 64q^2\ +\ 48q\ +\ 8\ +\ 1$ 

$a^2= 8(8q^2 + 6q +1) + 1$ 

$a^2\ =\ 8m\ +\ 1$  

Where, $m\ =\ 8q^2 +\ 6q\ +\ 1$

Taking, $r\ =\ 5$, 

$a\ =\ (8q\ +\ 5)$ 

Squaring both sides:

$a^2\ =\ 64q^2\ +\ 80q\ +\ 25$

$a^2\ =\ 64q^2\ +\ 80q\ +\ 24\ +\ 1$ 

$a^2\ =\ 8(8q^2\ +\ 10q\ +\ 3)\ +\ 1$ 

$a^2\ =\ 8m\ +\ 1$  

Where, $m\ =\ 8q^2\ +\ 10q\ +\ 3$

Taking, $r\ =\ 7$, 

$a\ =\ 8q\ +\ 7$ 

Squaring both sides:

$a^2\ =\ (8q\ +\ 7)^2$  

$a^2\ =\ 64q^2\ +\ 112q\ +\ 49$

$a^2\ =\ 64q\ +\ 112q\ +\ 48\ +\ 1$

$a\ =\ 8(8q\ +\ 14q\ +\ 6)\ +\ 1$ 

$a^2\ =\ 8m\ +\ 1$  

Where, $m\ =\ 8q\ +\ 14q\ +\ 6$

Hence, the square of any odd positive integer is of the form $8q\ +\ 1$.

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Updated on: 10-Oct-2022

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