Sure, let's solve the problem step-by-step. We want to prove that the square of a number that is two more than a multiple of 3 is one more than a multiple of 3.
1. Denote the number:
Let the number be expressed as [tex]\(3n + 2\)[/tex] where [tex]\(n\)[/tex] is an integer.
2. Expand the square of the number:
[tex]\[
\begin{align*}
(3n + 2)^2 & = (3n)^2 + 2 \cdot (3n) \cdot 2 + 2^2 \\
& = 9n^2 + 12n + 4
\end{align*}
\][/tex]
3. Rewrite the expression to isolate a multiple of 3:
[tex]\[
9n^2 + 12n + 4
\][/tex]
We want to express it as a sum of a multiple of 3 and an additional number. Notice that:
[tex]\[
9n^2 + 12n + 4 = 9n^2 + 12n + 3 + 1
\][/tex]
Group terms to show the multiple of 3 explicitly:
[tex]\[
= (9n^2 + 12n + 3) + 1
\][/tex]
4. Factor out the common term:
[tex]\[
9n^2 + 12n + 3
\][/tex]
can be factored as:
[tex]\[
3 \cdot (3n^2 + 4n + 1)
\][/tex]
Therefore, the expression becomes:
[tex]\[
= 3 \cdot (3n^2 + 4n + 1) + 1
\][/tex]
5. Conclusion:
[tex]\[
9n^2 + 12n + 4 = 3 \cdot (3n^2 + 4n + 1) + 1
\][/tex]
The term [tex]\(3 \cdot (3n^2 + 4n + 1)\)[/tex] is clearly a multiple of 3, and we have an additional [tex]\(1\)[/tex]. Thus, it is proven that the square of a number that is two more than a multiple of 3 is one more than a multiple of 3.
So, the complete step-by-step proof is given above. This demonstrates that [tex]\((3n + 2)^2\)[/tex] is indeed always one more than a multiple of 3.
