To prove that \( 2^{1403} + 1 \) is divisible by 3, we can use properties of modular arithmetic, specifically focusing on the behavior of powers of 2 modulo 3.

### Step 1: Understand the Pattern of \( 2^n \mod 3 \)
First, we need to investigate the pattern that the powers of 2 follow when taken modulo 3.

Calculate the first few powers of 2 modulo 3:

\[
\begin{aligned}
2^1 &\equiv 2 \pmod{3}, \\
2^2 &\equiv 4 \equiv 1 \pmod{3}, \\
2^3 &\equiv 2 \times 2^2 \equiv 2 \times 1 \equiv 2 \pmod{3}, \\
2^4 &\equiv 2 \times 2^3 \equiv 2 \times 2 \equiv 4 \equiv 1 \pmod{3}.
\end{aligned}
\]

We observe that the powers of 2 modulo 3 follow a repeating cycle:

\[
\begin{aligned}
2^1 &\equiv 2 \pmod{3}, \\
2^2 &\equiv 1 \pmod{3}, \\
2^3 &\equiv 2 \pmod{3}, \\
2^4 &\equiv 1 \pmod{3}, \\
\text{and so on.}
\end{aligned}
\]

The pattern is \( 2, 1, 2, 1, \dots \).

### Step 2: Determine the Exponent \( 1403 \mod 2 \)
The next step is to determine the position of \( 2^{1403} \) in this cycle. Notice that the cycle length is 2. Therefore:

\[
1403 \mod 2 = 1
\]

This tells us that:

\[
2^{1403} \equiv 2^1 \equiv 2 \pmod{3}
\]

### Step 3: Prove Divisibility
Now, substitute \( 2^{1403} \equiv 2 \) into the expression \( 2^{1403} + 1 \):

\[
2^{1403} + 1 \equiv 2 + 1 \equiv 3 \equiv 0 \pmod{3}
\]

### Conclusion
Since \( 2^{1403} + 1 \equiv 0 \pmod{3} \), \( 2^{1403} + 1 \) is indeed divisible by 3.

If you have any questions or would like to explore more details, feel free to ask!

Here are 5 related questions you might explore:

1. How can you prove that \( 2^{2n} - 1 \) is divisible by 3 for any integer \( n \)?
2. What other patterns do powers of integers modulo \( n \) exhibit?
3. Can you generalize this approach to prove divisibility for expressions like \( 5^{1403} + 2 \)?
4. What are some applications of modular arithmetic in cryptography?
5. How does Fermat's Little Theorem relate to this problem?

**Tip:** Identifying repeating cycles in modular arithmetic is a powerful technique that can simplify complex problems significantly.

Learn how to prove that 2^1403 + 1 is divisible by 3 using properties of modular arithmetic. This problem explores powers modulo 3 and the theorem of modular arithmetic. Suitable for advanced high school students and anyone interested in number theory.

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