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

Given: 

"Square of any odd integer is of the form $4m\ +\ 1$, for some integer m".

To do: 

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, $b\ =\ 4$. Then,

$a\ =\ 4q\ +\ r$, where $0 \underline{< } r < 4$.

When $r=0$,

$a=4 q$

$4 q$ is divisible by 2.

This implies, $4q$ is even.

When $r=1$,

$a=4 q+1$

$(4 q+1)$ is not divisible by 2 .
When $r=2$

$a=4 q+2$

$=2(2 q+1)$ which is divisible by 2.

This implies,

$2(2 q+1)$ is even.

When $r=3$,

$a=4 q+3$

$(4 q+3)$ is not divisible by 2 .
Therefore, for any positive integer $q,(4 q+1)$ and $(4 q+3)$ are odd integers.

$a^{2}=(4 q+1)^{2}$

$=16 q^{2}+1+8 q$

$=4(4 q^{2}+2 q)+1$ is a square which is of the form $4 m+1$, where $m=(4 q^{2}+2 q)$ is an integer.

$a^{2}=(4 q+3)^{2}$

$=16 q^{2}+9+24 q$

$=4(4 q^{2}+6 q+2)+1$ is a square which is of the form $4 m+1$, where $m=(4 q^{2}+6 q+2)$ is an integer.
Hence, for some integer $m$, the square of any odd integer is of the form $4 m+1$.

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Updated on: 10-Oct-2022

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