Question

Prove that the cube of any integer a has to be exactly one of these forms: 9K+1 or 9K+8 for some integerk.

Short answer

It is proved that the cube of any integer has to be exactly one of these forms: $9k+1$ or $9k+8$ for some integer.

Step-by-step solution

The Division Algorithm

Theorem 1.1 states that consider $a,b$ as an integer with $b>0$ . Then $q$ and$r$ are unique integersin which $a=bq+r$ and$0\le r\le b$ .

Theorem 1.1 permits for a negative dividend $a$; however, the remainder $r$ should not only be less than the divisor $b$, and it also should be nonnegative.

Show that the cube of any integer a   has to be exactly one of these forms:   9k+1or 9k+8 for some integer k

It is known that any integer $a$is of the form $3q,3q+1$or $3q+2$ for each $q$.

Taking cube on both sides of the equation $a=3q$ as follows:

$\begin{array}{rcl}{a}^3& =& {\left(3q\right)}^3\\ & =& 27q^3\\ & =& 9\left(3q^3\right)\end{array}$

It is observed that $a=3q$ for any integer $q$by letting $k=3q^3$.

Taking cube on both sides of the equation $a=3q+1$ as follows:

localid="1652785788496" $\begin{array}{rcl}{a}^3& =& {\left(3q+1\right)}^3\\ & =& 27q^3+27q^2+9q+1\\ & =& 9\left(3q^3+3q^2+q\right)+1\end{array}$

It is observed that $a^3=9k+1$ for any integer $k$ by letting localid="1646050925265" $k=3q^3+3q^2+q$.

Taking cube on both sides of the equation $a=3q+2$as follows:

localid="1652785840862" $\begin{array}{rcl}{a}^3& =& {\left(3q+2\right)}^3\\ & =& 27q^3+54q^2+36q+8\\ & =& 9\left(3q^3+6q^2+4q\right)+8\end{array}$

It is observed that $a^2=9k+8$for some integers by letting$k=3q^3+6q^2+4q$.

Hence, it is proved that the cube of any integer has to be exactly one of these forms: $9k+1$ or $9k+8$for some integer.