Show that square of any odd positive integer is in form 4m+1 for some

1.	Show that square of any odd positive integer is in form 4m+1 for some integer m
Answer» Let a be any odd positive integer, then on dividing a by b, we have\xa0a = bq+r,{tex}\u200b\u200b\u200b\u200b0 \\leqslant r < b{/tex}.....(i) [by Euclid\'s division lemma]On putting b = 2 in Eq.(i), we get\xa0a = 2q+r,\xa0{tex}0\\leqslant r < 2{/tex}\xa0⇒ r = 0 or 1If 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\xa0= (2q+1)2\xa0\xa0= 4q2\xa0+ 1 + 4q {tex}\\lbrack\\therefore(x+y)^2=x^2+y^2+2xy\\rbrack{/tex}= 4( q2\xa0+ q ) + 1\xa0= 4m + 1, where m = q2\xa0+ qHence, for some integer m, the square of any odd integer is of the form 4m+1.Hence Proved.

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