**Question:**

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

**Solution:**

By Euclid’s division algorithm, we have a = bq + r, where 0

On putting b = 4 in Eq. (i), we get

$a=4 q+r_{1}$ where $0 \leq r<4$ i.e., $r=0,1,2,3 \quad$...(ii)

If $r=0 \Rightarrow a=4 q, 4 q$ is divisible by $2 \Rightarrow 4 q$ is even.

If $r=1 \Rightarrow a=4 q+1,(4 q+1)$ is not divisible by 2 .

If $r=2 \Rightarrow a=4 q+2,2(2 q+1)$ is divisible by $2 \Rightarrow 2(2 q+1)$ is even.

If $r=3 \Rightarrow a=4 a+3(4 q+3)$ is not divisible by 2

So, for any positive integer $q,(4 q+1)$ and $(4 q+3)$ are odd integers.

Now, $\quad a^{2}=(4 q+1)^{2}=16 q^{2}+1+8 q=4\left(4 q^{2}+2 q\right)+1$

$\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$

is a square which is of the form $4 m+1$, where $m=\left(4 q^{2}+2 q\right)$ is an integer. and $a^{2}=(4 q+3)^{2}=16 q^{2}+9+24 q=4\left(4 q^{2}+6 q+2\right)+1$ is a square

$\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$

which is of the form 4m + 1, where m = (4q2 + 6q + 2) is an integer.

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

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