Chapter 8.4, Problem 57PPE

(a)

To determine

To prove: If an integer is one more than a multiple of 3 , then its square is also one more than a multiple of 3.

Explanation of Solution

Given information:

The integer is $3m+1$

Calculation:

  ${(3m+1)}^2$

  ${(a+b)}^2=a^2+b^2+2ab$

  ${(3m+1)}^2=9m^2+1+6m$

  ${(3m+1)}^2=9m^2+6m+1$

As $9m^{2}and\text{ 6m}$ are multiple of 3 so there sum is also multiple of 3

Hence its square is also one more than a multiple of 3.

Conclusion:

If an integer is one more than a multiple of 3 , then its square is also one more than a multiple of 3.

(b)

To determine

To prove: If an integer is two more than a multiple of 3 , then its square is not two more than a multiple of 3.

Explanation of Solution

Given information:

  $3m+2$

Calculation:

The integer $3m+2$.

So the square of $3m+2$ will be:

  ${(3m+2)}^2$

  ${(a+b)}^2=a^2+b^2+2ab$

  ${(3m+2)}^2=9m^2+4+12m$

  ${(3m+2)}^2=9m^2+12m+4$

As $9m^{2}and\text{ 12m}$ are multiple of 3 so there sum is also multiple of 3

But now it is 4 more than the multiple of 3

So its square is not two more than a multiple of 3.

Conclusion:

If an integer is two more than a multiple of 3 , then its square is not two more than a multiple of 3.