Question

Show that the square of an odd positive integer is of the form 8m+1, where m is some whole number.​

Answer 1

Answer:

see the image

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Answer 2

★To prove:-

The square of an odd positive integer is of the form 8m+1, where 'm' is some whole number.

★Proof:-

We know:

Euclid's division lemma:-

If a & b are two positive integers, the whole numbers q and r satisfy the equation: a = bq + r, where 0 ≤ r < b.

→a is the dividend

→ b is the divisor.

Here, b = 8

→ r ∈ [1, 7 ] i.e, r = 1, 2, 3,.....7

Hence,

→ a = 8q + r

Case - (1):-

Taking r = 1,

⇒ a = 8q + 1

Squaring on both sides,

$\displaystyle \longrightarrow \sf a^2 = (8q+1)^2\\\\\longrightarrow \sf a^2 = 64q^2+16q+1\\\\\longrightarrow \sf a^2 = 8(8q^2 +2q)+1\\\\\longrightarrow \sf a^2 = \bold{8m + 1}$

Where, m = 8q²+2q

Case - (2):-

Taking r = 3,

⇒ a = 8q + 3

Squaring on both sides,

$\displaystyle \longrightarrow \sf a^2 = (8q+3)^2\\\\\longrightarrow \sf a^2 = 64q^2 + 9+48q\\\\\longrightarrow \sf a^2 = 64q^2+48q + 8 +1\\\\\longrightarrow \sf a^2 = 8(8q^2 + 6q +1 )+1\\\\\longrightarrow \sf a^2 = \bold{8m+1}$

Where,m = 8q²+6q + 1

Case - (3):-

Taking r = 5,

⇒ a = 8q + 5

Squaring on both sides,

$\displaystyle \longrightarrow \sf a^2 = (8q+5)^2\\\\\longrightarrow \sf a^2 = 64q^2+80q+25\\\\\longrightarrow \sf a^2 = 64q^2 + 80q + 24 + 1\\\\\longrightarrow \sf a^2 = 8(8q^2 + 10q +3)+1 \\\\\longrightarrow \sf a^2 = \bold{8m+1}$

Where,m = 8q²+10q+3

We can see that the square of an odd positive integer is of the form 8m+1.

Hence proved !

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