Proving Divisibility: Understanding the Case (3^{2014} - 3^{2016}) by 15

Introduction

This article delves into the mathematical proof of the divisibility of (3^{2014} - 3^{2016}) by 15. It serves as a step-by-step guide to understanding the concept using methods from Modular Arithmetic and Proof by Factors.

Modular Arithmetic

Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value, known as the modulus. In this context, the modulus is 15, and we will use it to prove the divisibility of (3^{2014} - 3^{2016}).

Simplifying the Expression

The expression can be simplified as follows:

$$3^{2014} - 3^{2016} 3^{2014}(1 - 3^2) 3^{2014} cdot (1 - 9) 3^{2014} cdot (-8) -3^{2014} cdot 8$$

However, we want to show that the original expression (3^{2014} - 3^{2016}) is divisible by 15. Since (15 3 cdot 5), we will check for divisibility by these prime factors separately.

Proof by Factors

Step 1: Check Divisibility by 3

First, observe that (3^{2014}) is clearly divisible by 3, as it is a power of 3. Therefore, (3^{2014} - 3^{2016}) is also divisible by 3.

Step 2: Check Divisibility by 5

Next, we need to check if (3^{2014} cdot 10) is divisible by 5. Notice that (10 2 cdot 5), and hence, it is divisible by 5. Consequently, (3^{2014} cdot 10) is also divisible by 5.

Conclusion

Since (3^{2014} - 3^{2016}) is divisible by both 3 and 5, it follows that it is divisible by 15.

Alternate Methods

Method 1: Mod Simplification

We can also use simplification of modulo operations to prove the result. Let (S 3^{2014} cdot 3^2 - 1). Then:

$$S mod 15 3^{2014} cdot 9 - 1 mod 15 10 cdot 3^{2014} mod 15$$

Now, note that 15 3 times 5, and the two factors are coprime. We compute S mod 3 and S mod 5 separately.

Mod 3 Computation:

$$S mod 3 100 mod 3 0 mod 3$$

Mod 5 Computation:

$$S mod 5 0 cdot 3^{2014} mod 5 0 mod 5$$

Since 3 and 5 are coprime, it follows immediately that S mod 15 0 mod 15.

Method 2: Periodicity of Mod 15

Another approach involves observing the periodicity of 3 mod 15. We have:

$$ 3^1 equiv 3 pmod{15} 3^2 equiv 9 pmod{15} 3^3 equiv 12 pmod{15} 3^4 equiv 6 pmod{15} 3^5 equiv 3 pmod{15} $$

Thus, 3 has a periodicity of 4. Therefore, 3^{2014} times 3^{2016} equiv 9 times 6 equiv 0 pmod{15}.

Conclusion

In summary, (3^{2014} - 3^{2016}) is divisible by 15, as we have shown through different methods: divisible factors, modular arithmetic, and periodicity.