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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 Lemma
a=bq+r

Here a is any positive integer & b=4

& 0r<4r=0,1,2,3

a=4 q+r.....(1)

Put r=0 in equation (1)

a=4 q

Cubing on both sides

(a)3=(4 q)3

(a)3=64 q3=4(16 q3)

a3=4 m where m=16 q3

Put =1 in equation (1)

a=4 q+1 cubing both side

a3=(4 q+1)3

we know

(a+b)3=a3+b3+3a3b+3a b2

so apply

a3=(4 q)3+(1)3+3(4 q)2×1+3(4 q)×(1)2

a3=64 q3+1+3×16 q2+12 q

a3=64 q3+1+48 q2+12 q

a3=4(16 q3+12 q3+3 q)+1

a3=4 m+1

where m=16 q3+12 q2+3 q

again put r=2 in equation (1)

a=4 q+2

cube both sides

a3=(4 q+2)3

a3=(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×4

a3=64 q3+8+96 q2+48 q

a3=4(16 q3+2+24 q3+12 q)

a3=4 m

m=16 q3+2+24 q3+12 q

Put r=3 in equation (1)

a=4 q+3

again cube

a3=(4 q+3)3

a3=(64 q3+27+144 q2+108 q)+3

a3=(64 q3+24+144 q2+108 q)+3

a3=4 m+3

when m=16 q3+6+36 q2+27

hence a3 is of the for
4 m, 4 m+1,4 m+3

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