Show that square of any odd …
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Posted by Rohan Vaishnav 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let a be any odd positive integer, then on dividing a by b, we have
a = bq+r,{tex}0 \leqslant r < b{/tex}.....(i) [by Euclid's division lemma]
On putting b = 2 in Eq.(i), we get
a = 2q+r, {tex}0\leqslant r < 2{/tex}
⇒ r = 0 or 1
If r = 0, then a = 2q, which is divisible by 2. So, 2q is even.
If {tex}r = 1{/tex}, then {tex}a = 2q + 1{/tex}, which is not divisible by 2. So {tex}(2q+1){/tex} is odd.
Now, as a is odd, so it cannot be of the form 2q.
Thus, any odd positive integer a is of the form (2q+1).
Now, consider a2 = (2q+1)2 = 4q2 + 1 + 4q {tex}\lbrack\therefore(x+y)^2=x^2+y^2+2xy\rbrack{/tex}
= 4( q2 + q ) + 1 = 4m + 1, where m = q2 + q
Hence, for some integer m, the square of any odd integer is of the form 4m+1.
Hence Proved.
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