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

1.	Show that square of any odd integer is of the form 4m+1 for some integer m.
Answer» Let a be any positive integer and b = 4\xa0Then, by Euclid\'\'s algorithm a = 4q + r for some integer q\xa00 and 0\xa0\xa0r < 4\xa0Since, r = 0, 1, 2, 3\xa0Therefore, a = 4q or 4q + 1 or 4q + 2 or 4q + 3\xa0Since, a is an odd integer, o a = 4q + 1 or 4q + 3\xa0Case I:\xa0When a = 4q + 1\xa0Squaring both sides, we have,\xa0a2\xa0= (4q + 1)2\xa0\xa0a2\xa0= 16q2\xa0+ 1 + 8q\xa0 = 4(4q2\xa0+ 2q) + 1\xa0 = 4m + 1 where m = 4q2\xa0+ 2q Case II:\xa0When a = 4q + 3Squaring both sides, we have,a2\xa0= (4q +3)2 = 16q2\xa0+ 9 + 24q = 16 q2\xa0+ 24q + 8 + 1 = 4(4q2\xa0+ 6q + 2) +1 = 4m +1 where m = 4q2\xa0+7q + 2Hence, a is of the form 4m + 1 for some integer m.\xa0 Refer ncert

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