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Question

# 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

## By euclids division Lemmaa=bq+rHere a is any positive integer & b=4& 0≤r<4r=0,1,2,3a=4 q+r.....(1)Put r=0 in equation (1)a=4 qCubing on both sides(a)3=(4 q)3(a)3=64 q3=4(16 q3)a3=4 m where m=16 q3Put =1 in equation (1)a=4 q+1 cubing both sidea3=(4 q+1)3we know(a+b)3=a3+b3+3a3b+3a b2so applya3=(4 q)3+(1)3+3(4 q)2×1+3(4 q)×(1)2a3=64 q3+1+3×16 q2+12 qa3=64 q3+1+48 q2+12 qa3=4(16 q3+12 q3+3 q)+1a3=4 m+1where m=16 q3+12 q2+3 qagain put r=2 in equation (1)a=4 q+2cube both sidesa3=(4 q+2)3a3=(4 q)3+(2)3+3×(4 q)2×2+3×4 q(2)a3=64 q3+8+3×16 q2×2+3×4 q×4a3=64 q3+8+96 q2+48 qa3=4(16 q3+2+24 q3+12 q)a3=4 mm=16 q3+2+24 q3+12 qPut r=3 in equation (1)a=4 q+3again cubea3=(4 q+3)3a3=(64 q3+27+144 q2+108 q)+3a3=(64 q3+24+144 q2+108 q)+3a3=4 m+3when m=16 q3+6+36 q2+27hence a3 is of the for4 m, 4 m+1,4 m+3

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