Question

Question 5 Show that the square of any odd integer is of the form 4m + 1, for some integer m.

Solution

By Euclid's division algorithm, we have a=bq+r, where 0≤r<b On putting b = 4, we get, a=4q+r where 0≤r<4 i.e., r = 0, 1, 2, 3 If r = 0 ⇒  a = 4q is divisible by 2  ⇒  4q is even. If r = 2⇒ a=4q+2,(22q+1) is divisible b y 2 ⇒  2 (2q + 1) is even. If r = 3⇒ a=4q+3,(4q+1) and (4q+3) are odd integers. So, for any positive integer q, (4q + 1) and (4q + 3) are odd integers. Now,  a2=(4q+1)2=16q2+1+8q=4(4q2+2q)+1 [∵ (a+b)2=a2+2ab+b2] This is a square which is of the form 4m + 1, where m = (4q2+2q) is an integer. and a2=(4q+3)2=16q2+9+24q=4(4q2+6q+2)+1 is a square. [∵ (a+b)2=a2+2ab+b2] Which is of the form 4m + 1, where m =  (4q2+6+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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