Name: Fundamental Theorem of Arithmetic
Uploaded: 2019-07-19
Duration: 5 min 4 s
Description: A number can be represented in the form of the Product of their Primes. This method is also called the Fundamental Theorem of Arithmetic. Watch the video to...

Question

Show that the cube of a positive integer of the form $6 q + r, q$ is an integer and $r = 0, 1, 2, 3, 4, 5$ is also of the form $6 m + r .$

Accepted Answer

Let $a$ be an arbitrary positive integer. Then, by Euclid's division algorithm, corresponding to the positive integers $a$ and $6$ , there exist non-negative integers $q$ and $r$ such that

$a = 6 q + r,$ where, $0 \leq r < 6$

$\Rightarrow a^{3} = {(6 q + r)}^{3} = 216 q^{3} + r^{3} + 3.6 q . r (6 q + r)$

$[∵ {(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b)]$

$\Rightarrow a^{3} = (216 q^{3} + 108 q^{2} r + 18 q r^{2}) + r^{3} . .. (i)$

where, $0 \leq r < 6$

Case $I$ When $r = 0$ , then putting $r = 0$ in Eq. $(i)$ , we get

$a^{3} = 216 q^{3} = 6 (36 q^{3}) = 6 m$

where, $m = 36 q^{3}$ is an integer.

Case $I I$ When $r = 1$ , then putting $r = 1$ in Eq. $(i)$ , we get

$a^{3} = (216 q^{3} + 108 q^{3} + 18 q) + 1 = 6 (36 q^{3} + 18 q^{3} + 3 q) + 1$

$\Rightarrow a^{3} = 6 m + 1$ , where $m = (36 q^{3} + 18 q^{3} + 3 q)$ is an integer.

Case $I I I$ When $r = 2$ , then putting $r = 2$ in Eq. $(i)$ , we get

$a^{3} = (216 q^{3} + 216 q^{2} + 72 q) + 8$

$a^{3} = (216 q^{3} + 216 q^{2} + 72 q + 6) + 2$

$\Rightarrow a^{3} = 6 (36 q^{3} + 36 q^{2} + 12 q + 1) + 2 = 6 m + 2$

where, $m = (36 q^{3} + 36 q^{2} + 12 q + 1)$ is an integer.

Case $I V$ When $r = 3$ , then putting $r = 3$ in Eq. $(i)$ , we get

$a^{3} = (216 q^{3} + 324 q^{2} + 162 q) + 27 = (216 q^{3} + 324 q^{2} + 162 q + 24) + 3$

$= 6 (36 q^{3} + 54 q^{2} + 27 q + 4) + 3 = 6 m + 3$

where, $m = (36 q^{2} + 54 q^{2} + 27 q + 4)$ is an integer.

Case $V$ When $r = 4$ , then putting $r = 4$ in Eq. $(i)$ , we get

$a^{3} = (216 q^{2} + 432 q^{2} + 288 q) + 64$

$= 6 (36 q^{3} + 72 q^{2} + 48 q) + 60 + 4$

$= 6 (36 q^{3} + 72 q^{2} + 48 q + 10) + 4 = 6 m + 4$

where, $m = (36 q^{3} + 72 q^{2} + 48 q + 10)$ is an integer.

Case $V I$ When $r = 5$ , then putting $r = 5$ in Eq. $(i)$ , we get

$a^{3} = (216 q^{3} + 540 q^{2} + 450 q) + 125$

$\Rightarrow a^{3} = (216 q^{3} + 540 q^{2} + 450 q) + 120 + 5$

$\Rightarrow a^{3} = 6 (36 q^{3} + 90 q^{2} + 75 q + 20) + 5$

$\Rightarrow a^{3} = 6 m + 5$

where, $m = (36 q^{3} + 90 q^{2} + 75 q + 20)$ is an integer.

Hence, the cube of a positive integer of the form $6 q + r$ , where $q$ is an integer and $r = 0, 1, 2, 3, 4, 5$ is of the forms $6 m, 6 m + 1, 6 m + 2, 6 m + 3, 6 m + 4$ and $6 m + 5$ i.e., $6 m + r$ .

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