Question

Show that the square of an odd integer is expressible in the form 3k or 3k +1. Do by own.

Answer 1

Well, you know that any integer can be written in the form , or following the division algorithm. 3k 3k + 1 3k + 2
Now let's take the square of each of them;
(3k) = 9 = 3(3 ),
of the form ;
 2 k 2 k 2 3k (3k + 1) = 9 + 6k + 1 = 3(3 + 2k) + 1,
 of the form ;
2 k 2 k 2 3k + 1 (3k + 2) = 9 + 12k + 4 = 9 + 12k + 3 + 1 = 3(3 + 4k + 1) + 1,
 of the form ;
2 k 2 k 2 k 2 3k + 1 therefore no integer can be written in the form ,
 hence all integers squared are of the form ,
 or . n 2 3k + 2 3k 3k + 1......


hope it helps u...... ;)

@Av.....

Answer 2

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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