Show that the square of an odd integer is expressible in the form 3k or 3k +1. Do by own.
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According to euclids division lemma ,we get
a=bq+ r, where 0<=rLet us take b as 3 ,then r=0,1,2.
Now substituting the values,
If r=0, then a=3q
If r =1,then a=3q +1
If r=2,then a=3q+2
Now a²= (3q)²= 9q²=3(3q²)=3k where k=3q².
a²=(3q+1)²=9q²+6+1=3(3q²+2) +1=3k+1 where k=3q²+2.
a²=(3q+2)²=9q²+12+4=9q²+12+3+1= 3(3q²+4+1)+1=3(3q²+5)+1=3k +1 where k = 3q²+5.
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a=bq+ r, where 0<=rLet us take b as 3 ,then r=0,1,2.
Now substituting the values,
If r=0, then a=3q
If r =1,then a=3q +1
If r=2,then a=3q+2
Now a²= (3q)²= 9q²=3(3q²)=3k where k=3q².
a²=(3q+1)²=9q²+6+1=3(3q²+2) +1=3k+1 where k=3q²+2.
a²=(3q+2)²=9q²+12+4=9q²+12+3+1= 3(3q²+4+1)+1=3(3q²+5)+1=3k +1 where k = 3q²+5.
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