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Proving 7^n - 4^n is Divisible by 3 Using Mathematical Induction and Modular Arithmetic

April 11, 2025Technology2681
Proving 7^n - 4^n is Divisible by 3 Using Mathematical Induction and M

Proving 7^n - 4^n is Divisible by 3 Using Mathematical Induction and Modular Arithmetic

This article delves into the methods of proving that 7n - 4n is divisible by 3 for all positive integers n. We explore two methods: mathematical induction and modular arithmetic. Both methods provide rigorous and elegant proofs of the statement.

Mathematical Induction Approach

One way to prove that 7n - 4n is divisible by 3 for all integers n is through the principle of mathematical induction.

Base Case

Consider the base case when n 1:

71 - 41 7 - 4 3, which is divisible by 3. Therefore, the base case holds.

Inductive Step

Assume that the statement is true for some integer k, i.e., 7k - 4k is divisible by 3. This means:

7k - 4k ≡ 0 (mod 3)

Our goal is to show that 7(k 1) - 4(k 1) is also divisible by 3. Consider the expression:

7(k 1) - 4(k 1) 7k ? 7 - 4k ? 4 7k ? 7 - 4k ? 4 - 4k ? 7 4k ? 7

Factoring out common terms, we get:

7(k 1) - 4(k 1) (7k - 4k) ? 7 - 4k ? 3

Since we assumed 7k - 4k ≡ 0 (mod 3), we know that (7k - 4k) ? 7 ≡ 0 (mod 3) and 4k ? 3 ≡ 0 (mod 3). Therefore:

7(k 1) - 4(k 1) ≡ 3 ? 7k - 4k ? 3 ≡ 0 (mod 3)

Modular Arithmetic Approach

An alternative approach involves modular arithmetic. We can express 7n - 4n in modular notation and demonstrate its divisibility by 3.

Using Polynomial Factorization

Consider the expression 7n - 4n. This resembles the polynomial structure xn - yn, which can be factored as:

xn - yn (x - y) ? (x(n-1) x(n-2) ? y x(n-3) ? y2 … x ? y(n-2) y(n-1))

Substitute x 7 and y 4:

7n - 4n (7 - 4) ? (7(n-1) 7(n-2) ? 4 7(n-3) ? 42 … 7 ? 4(n-2) 4(n-1))

Since 7 - 4 3, we have:

7n - 4n ≡ 3 ? (7(n-1) 7(n-2) ? 4 7(n-3) ? 42 … 7 ? 4(n-2) 4(n-1)) ≡ 0 (mod 3)

Using Congruences

Another approach involves modular congruences:

7 ≡ 1 (mod 3) rarr; 7n ≡ 1n 1 (mod 3)

4 ≡ 1 (mod 3) rarr; 4n ≡ 1n 1 (mod 3)

Therefore, subtracting 7n by 4n modulo 3:

7n - 4n ≡ 1 - 1 0 (mod 3)

Conclusion

Using both mathematical induction and modular arithmetic, we have rigorously proven that 7n - 4n is divisible by 3 for all positive integers n.

Understanding these methods not only confirms the mathematical claim but also provides insights into the underlying principles of number theory and algebra. These techniques are fundamental in both theoretical and applied mathematics.