Prove by mathematical induction
\[
9^{n} + 3
\]
is divisible by 4 for all natural numbers \( n \gt 0 \).
Prove the statement is true for a particular n. Here n = 1 is a reasonable choice.
To show this is true for n=1, observe that the expression = 12. Now 12 is divisible by 4, so the statement is true for n = 1.
Assume the statement is true for n = k and show it is true for n = k + 1.
To show the statement is true for n = k+1,
\[ \begin{align}
9^{(k+1)} + 3 &= 9 \times 9^k + 3 \\
\end{align} \ \]
Use the inductive hypothesis to simplify the right hand side.
By the inductive hypothesis,
\[ \begin{align}
9 \times 9^{k} + 3 &= 9 \times (4m-3) + 3 \quad \text{for some integer m}\\
&= 9 \times 4 m - 27 + 3 \\
&= 4 \times 9m - 24 \\
&= 4 \times (9m - 6)
\end{align} \ \]
Which is divisible by 4. So the statement is true for n = k+1.
By the principle of mathematical induction,
The statement is true for \( n \gt 0 \).
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