Question 4:
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m.
Solution:
Let us assume that x be any positive integer and take y=3.
Then by applying Euclid division algorithm we have
X = 3q+r, for integer q >=0 and r=0,1,2.
By putting the values we can have
X =3q, 3q+1,3q+2.
Now by squaring both the side we can have
X^2 =9q^2 = 3*3q^2.
Let 3q^2=m.
Therefore we have X^2= 3m………………. (1)
And x^2= (3q+1)^2 = 9q^2 +1+6q = 3(3q^2+2)+1.
Substituting 3q^2+2q =m, we get
X^2 = 3m+1…………………..(2)
Now we have,
X^2 =(3q+2)^2= 9q^2+4+12q = 3 (3q^2+4q+1)+1.
Again substitute the value of 3q^2+4q+1= m, we get
X^2 = 3m+1 ………………..(3)
Hence from above equations we can say that square of any positive integer is either of the form 3m or 3m+1.
Question 5:
Use Euclid division lemma to prove that cube of any positive integer is of the form 9m, 9m+1 or 9m+8.
Solution :
Let us assume that x is any positive integer and y=3.
Now applying Euclid division algorithm we get
X =3q+r, where q>=0 and r=0,1,2.
Now putting the values of r we will get the values of x as
X=3q
X=3q+1
X=3q+2
Now by taking the cube of above expression we have some values of x as
Case 1 : when r = 0
X^3 = 27q^3= 9 (3q^3)= 9m. Where m = 3q^3.
Case 2 : when r =1
X ^3 =(3q+1)^3= 27q ^3+1+27q^2+9q.
Solving the above equation by taking 9 as common factor, we get
X^3 = 9(3q^3+3q^2+q)+1.
Now putting m = 3q^3+3q^2+q, we will have
X^3 =9m+1.
Case 3 : when r =2
X^3 = (3q+2)^3 = 27q^3 +54q^2+36q +8.
Now taking 9 as common factor in above equation we get
X ^3 =9m+8 , where m = 3q^3+6q^2+4q.
Hence from above cases it is proved that cube of any positive integer is of the form 9m, 9m+1 or 9m+8.
