Question

show that the cube of any positive integer is of the form 4m or 4m+1 or 4m+3 where‘m’ is a whole number​

Answer 1

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.

(4q)3 = 64q3 = 4(16q3) = 4m, where m is some integer.

(4q + 1)3 = 64q3 + 48q2 + 12q + 1 = 4(16q3 + 12q2 + 3) + 1 = 4m + 1, where m is some integer.

(4q + 2)3 = 64q3 + 96q2 + 48q + 8 = 4(16q3 + 24q2 + 12q + 2) = 4m, where m is some integer.

(4q + 3)3 = 64q3 + 144q2 + 108q + 27 = 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.

Answer 2

Step-by-step explanation:

let a be the positive integer a and b =4

Then by eculid's division 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+3

(4q)3= 64q 3=4(16q3)=4m,where m is a some integer

(4q+1)3=64q3+48q2+12q+1=4 (16q 3+12q 2+3)+1=4m+1 where m is some integer

(4q+2)3=64q+3+96q2+48q+8=

4 (16q3 + 24q2+12q+2)=4m,where m is some integer

(4q+3) 3= 64q3+144q 2+108q+27=4 (16q3+36q2+27q+6)+3=4m+3

Hence the cube of any positive integer is of the form 4m,4m+1 or 4m+3 some integer m