Show that cube of any positive integer is of form of 4 m 4m+1 4m+8

1.	Show that cube of any positive integer is of form of 4 m 4m+1 4m+8
Answer» Let a be the positive integer and b = 4.Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.{tex}(4q)^3\\;=\\;64q^3\\;=\\;4(16q^3){/tex}= 4m, where m is some integer.{tex}(4q+1)^3\\;=\\;64q^3+48q^2+12q+1=4(\\;16q^3+12q^2+3q)+1{/tex}= 4m + 1, where m is some integer.{tex}(4q+2)^3\\;=\\;64q^3+96q^2+48q+8=4(\\;16q^3+24q^2+12q+2){/tex}= 4m, where m is some integer.{tex}(4q+3)^3\\;=\\;64q^3+144q^2+108q+27{/tex}=4×(16q3+36q2+27q+6)+3= 4m + 3, where m is some integer.Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m. Thank you so much

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