Question

Show that if \(m\) is an integer then 3 divides \(m^{3}-m\).

Short answer

Question: Prove that for any integer m, 3 divides m^3 - m. Answer: To prove this statement, we consider the possible remainders when m is divided by 3, namely 0, 1, and 2. For each case, we show that m^3 - m is divisible by 3. Consequently, 3 divides m^3 - m for any integer m.

Step-by-step solution

Case 1: m is a multiple of 3

If \(m\) is a multiple of 3, then \(m = 3k\) for some integer \(k\). In this case, we can express \(m^3 - m\) as \((3k)^3 - 3k = (27k^3 - 3k)\). We can factor out a 3 from this expression to get \(3(9k^3 - k)\). Clearly, we see that this expression is divisible by 3.

Case 2: m has a remainder of 1 when divided by 3

If \(m\) has a remainder of 1 when divided by 3, then \(m = 3k + 1\) for some integer \(k\). In this case, we have \((3k + 1)^3 - (3k + 1) = (27k^3 + 27k^2 + 9k + 1) - (3k + 1) = 27k^3 + 27k^2 + 6k\). We can factor out 3 from the expression to get \(3(9k^3 + 9k^2 + 2k)\). This expression is divisible by 3.

Case 3: m has a remainder of 2 when divided by 3

If \(m\) has a remainder of 2 when divided by 3, then \(m = 3k + 2\) for some integer \(k\). In this case, we have \((3k + 2)^3 - (3k + 2) = (27k^3 + 54k^2 + 36k + 8) - (3k + 2) = 27k^3 + 54k^2 + 33k + 6\). We can factor out 3 from the expression to get \(3(9k^3 + 18k^2 + 11k + 2)\). This expression is divisible by 3. Since 3 divides \(m^3 - m\) for all possible remainders, we can conclude that if \(m\) is an integer, then 3 divides \(m^3 - m\).

Key concepts

Modular Arithmetic

Modular arithmetic is like clock arithmetic, focusing on the remainder after division. When we say "3 divides \(m^3 - m\)," we mean the remainder of this expression when divided by 3 is zero. It's about understanding numbers in terms of their remainders after division by another number, in this case, 3.

Consider different scenarios for integer \(m\):

• If \(m\) is a multiple of 3, like 3, 6, 9, etc., then \(m \equiv 0 \mod 3\). The expression \(m^3 - m\) simplifies to a form that is clearly divisible by 3.
• If \(m\) has a remainder of 1 when divided by 3, then \(m \equiv 1 \mod 3\). A little algebra shows that \((m^3 - m) \equiv 0 \mod 3\) again.
• If \(m\) has a remainder of 2 when divided by 3 (\(m \equiv 2 \mod 3\)), once more, \(m^3 - m\) becomes divisible by 3.

By proving all these cases, we show that no matter what integer \(m\) you pick, \(m^3 - m\) leaves no remainder when divided by 3.

Polynomial Factorization

Polynomial factorization involves breaking down a polynomial into simpler parts, usually products of polynomials of lower degree. It helps understand the properties of the polynomial, like roots or divisibility.

In this exercise, we express \(m^3 - m\) as a product of polynomials. Specifically, \(m^3 - m\) can be factored as \(m(m-1)(m+1)\). This factorization reveals that the expression has three consecutive integers as factors.

One of the important realizations with polynomial factorization is that among any three consecutive integers, one must be divisible by 3. Thus, either \(m\), \(m-1\), or \(m+1\) is divisible by 3, leading to the whole expression being divisible by 3. Factorization, therefore, is a powerful tool to confirm divisibility in this context.

Remainder Theorem

The Remainder Theorem provides a method to find the remainder of a polynomial divided by a linear divisor. If you have a polynomial \(P(x)\) and want the remainder when divided by \(x-a\), you can substitute \(a\) into \(P(x)\). The result is the remainder.

In simpler terms, for \(P(x) = x^3 - x\) and divisor \(x = 3\), checking \(P(3)\) helps verify divisibility. Here, it's done by recognizing that \(P(x)\) simplifies into consecutive factors. When evaluated at points like 3, the calculation confirms uniform divisibility across possible values of \(x\) over modulo constraints.

Utilizing the Remainder Theorem confirms that even under different cases or inputs (modulations like \(m = 3k, 3k+1, 3k+2\)), the polynomial \(m^3-m\) reliably outputs a remainder of zero. This theorem acts as a double-check to ensure mathematical accuracy beyond initial observation or factorization.