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# Question 2 Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some 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 4, there exist non-negative integers q and r such that a=4q+r,where0≤r<4⇒a3=(4q+r)3⇒64q3+r3+12qr2+48q2r⇒a3=(64q3+48q2r+12qr2)+r3 …(i) Where, 0≤r<4 Case I When r =0, Putting r = 0 in Equation (i), we get a3=64q3=4(16q3)=4m Where m=16q3 is an integer. Case II When r =1, Putting r = 1 in Equation(1), we get a3=64q3+48q2+12q+1=4(16q3+12q2+3q)+1=4m+1 Whre m=(16q3+12q2+3q) is an integer Case III When r = 2 Putting r = 2 Equation(1), we get a3=64q3+96q2+48q+8=4(16q3+24q2+12q+2)=4m Where m=(16q3+24q2+12q+2) is an integer. Case IV: When r =3. Putting r = 3 Equation(1), we get a3=64q3+144q2+108q+27=64q3+144q2+108q+24+3=4(16q3+36q2+27q+6)+3=4m+3 Where m=(16q3+36q2+27q+6) is an integer. Hence, the cube of any positive integer is of the form 4m, 4m + 1 or 4m +3 for some integer m.

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