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 Show that square of any positive odd integer is of the form 8m + 1, for some integer m.

Dear Student!

Here is the answer to your question.
 
Any odd positive integer is of the form 4q + 1 or 4q + 3 for some integer q.
When n = 4q + 1,

⇒ n2 is in the form 8m + 1
Hence, n2 is in the form 8m + 1 if n is an odd positive integer.
 
Cheers!

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View Full Answer

by euclids divison lemma for any 2 positive integers a and b there exits a unique interger q and r such that a=bq+r  ,, 0 is < or equal to r is greater than b

 

the possible values of r may be0,1,2,3,4,5,6,7.

if r=0 then a=8q+0

=a2=(8q)2

a2=64q2

  =8(8q2)

here 8q2 is m

if r=1 then a=8m+1

a2=(8q +1)2

  =64q2 + 16q+1

  = 8(8q2+2q)+1

here m= 8q2+2q

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