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# 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 0≤r<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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