Questions: Number Theory Prove: The square of a number that is two more than a multiple of 3 is one more than a multiple of 3. (3n+2)^2 = [?] n^2 + [?] n + [?] = [?] n^2 + [?] n + [?] + 1 = [?] ([?] n^2 + [?] n + [?]) + 1 = 1 more than a multiple of 3

Number Theory

Prove: The square of a number that is two more than a multiple of 3 is one more than a multiple of 3.

(3n+2)^2 = [?] n^2 + [?] n + [?] = [?] n^2 + [?] n + [?] + 1 = [?] ([?] n^2 + [?] n + [?]) + 1 = 1 more than a multiple of 3

Transcript text: Number Theory Prove: The square of a number that is two more than a multiple of 3 is one more than a multiple of 3 . \[ \begin{aligned} (3 \mathrm{n}+2)^{2} & \left.=[?] \mathrm{n}^{2}+\square \mathrm{n}+\square\right] \\ & \left.=\square \mathrm{n}^{2}+\square \mathrm{n}+\square\right]+1 \\ & =\square \quad\left(\square \mathrm{n}^{2}+\square \mathrm{n}+\square\right)+1 \\ & =1 \text { more than a multiple of } 3 \end{aligned} \]

Solution

Solution Steps

Step 1: Express the Number

Let \( n \) be an integer. We express the number as \( 3n + 2 \), which is two more than a multiple of 3.

Step 2: Square the Expression

We calculate the square of the expression: \[ (3n + 2)^2 = 9n^2 + 12n + 4 \]

Step 3: Simplify the Result

Next, we simplify the expression: \[ 9n^2 + 12n + 4 = 3(3n^2 + 4n) + 4 \]

Step 4: Analyze the Result

We can rewrite the expression as: \[ 3k + 1 \quad \text{where } k = 3n^2 + 4n + 1 \] This shows that \( 9n^2 + 12n + 4 \) is of the form \( 3k + 1 \), indicating that it is one more than a multiple of 3.

Step 5: Conclusion

Thus, we have proven that the square of a number that is two more than a multiple of 3 is indeed one more than a multiple of 3.