Question

show that cube of every positive integer is either of form 4m,4m+1,4m+2,4m+3​

Answer 1

Let n be any posi tivr no. then, it is of the form 4q,4q+1,4q+2 and 4q+3.

Case1: when n=4q

n^3=(4q)^3=(64q^3)=4m where m=16m^3

Case2: when n=4q+1

n^3=(4q+1)^3=64q^3+48q^2+12q+1=4(16q^3+12q^2+3q)+1=4m+1 where m= 16q^3+12q^2+3q

Case 3: When n=4q+2

n^3=(4q+2)^3=4(16q^3+24q^2+12q+2) where m=16q^3+24q^2+12q+2

Case 4: n=4q+3

n^3=(4q+3)^3=4(16q^3+36q^2+27q+6)+3=4m+3 where m=16q^3+36q^2+27q+6)

Chance, n is of the form 4m,4m+1,4m+2 and 4m+3.