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Question

Question 4
Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

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Solution

Let a be an arbitrary positive integer. Then by Euclid's division algorithm, corresponding to the positive integers a and 6, there exist non - negative integers q and r such that:
a=6q+r, where 0r<6
a2=(6q+r)2=36q2+r2+12qr [ (a+b)2=a2+2ab+b2]
a2=6(6q2+2qr)+r2 ...(i)

Case I
When r = 0, putting r = 0 in eq. (i), we get,
a2=6(6q2)=6m
Where, m=6q2 is an integer.

Case II
When r = 1, putting r = 1 in eq. (i) we get,
a2=6(6q2+2q)+1=6m+1
Where, m=(6q2+2q) is an integer.

Case III
When r = 2, putting r = 2 in eq (i), we get,
a2=6(6q2+4q)+4=6m+4
Where, m = 6q2+4q is an integer.

case IV
When r = 3, putting r = 3 in eq. (i) we get,
a2=6(6q2+6q)+9
=6(6q2+6q)+6+3
a2=6(6q2+6q+1)+3=6m+3
Where, m=(6q2+6q+1) is an integer.

Case V
When r = 4, putting r = 4 in eq. (i), we get,
a2=6(6q2+8q)+16
=6(6q2+8q)+12+4=6m+4
Where, m=(6q2+8q+2) is an integer.

Case VI
When r = 5, putting r = 5 in eq. (i), we get,
a2 = 6(6q2+10q)+25
= 6(6q2+10q)+24+1
a2=6(6q2+10q+4)+1=6m+1
Where, m=(6q2+10q+4) is an integer.
Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.


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