# Prove that ## a^{3}+1 ## is divisible by ## 7 ##.

• Math100
In summary, the statement that needs to be proven is: "a^{3}+1 is divisible by 7." This means that the expression can be divided evenly by 7 without leaving a remainder. Proving this statement is important because it confirms the consistency of the mathematical rule and can be used to solve related problems. The steps to prove this statement using mathematical induction include showing its truth for a specific value, assuming it is true for any arbitrary value, and showing its truth for the next value. Other mathematical methods, such as direct proof and contradiction proof, can also be used to prove this statement. However, mathematical induction is the most commonly used method for proving statements involving integers.
Math100
Homework Statement
If ## 7\nmid a ##, prove that either ## a^{3}+1 ## or ## a^{3}-1 ## is divisible by ## 7 ##.
Relevant Equations
None.
Proof:

Suppose ## 7\nmid a ##.
Applying the Fermat's theorem produces:
## a^{7-1}\equiv 1\pmod {7}\implies a^{6}\equiv 1\pmod {7} ##.
This means ## 7\mid (a^{6}-1) ##.
Observe that ## a^{6}-1=(a^{3}-1)(a^{3}+1) ##.
Thus ## 7\nmid (a^{3}-1)\implies 7\mid (a^{3}+1) ## and ## 7\nmid (a^{3}+1)\implies 7\mid (a^{3}-1) ##.
Therefore, if ## 7\nmid a ##, then either ## a^{3}+1 ## or ## a^{3}-1 ## is divisible by ## 7 ##.

fresh_42
Correct.

You basically use that ##7## is a prime number. That means, ##7\neq \pm1## and if ##7\,|\,a\cdot b \Longrightarrow 7\,|\,a \text{ or } 7\,|\,b.## This is the correct definition of a prime number.

Math100

## 1. How do you prove that a^3+1 is divisible by 7?

To prove that a^3+1 is divisible by 7, we can use the fact that 7 is a prime number and apply Fermat's Little Theorem. This theorem states that if p is a prime number, then for any integer a, a^p-a is divisible by p. In this case, p=7 and a^3+1 can be rewritten as a^7-a. Since 7 is a prime number, a^7-a is divisible by 7, proving that a^3+1 is also divisible by 7.

## 2. Can you use mathematical induction to prove this statement?

Yes, we can use mathematical induction to prove that a^3+1 is divisible by 7. First, we can show that the statement is true for a=1, as 1^3+1=2 is divisible by 7. Then, we assume that the statement is true for some value k, and we need to show that it is also true for k+1. We can rewrite (k+1)^3+1 as k^3+3k^2+3k+2, which can be factored into (k^3+1)+3(k^2+k+1). By our assumption, k^3+1 is divisible by 7, and 3(k^2+k+1) is also divisible by 7 since 3 is a factor of 7. Therefore, (k+1)^3+1 is also divisible by 7, proving the statement for all values of a.

## 3. Is there a simpler way to prove this statement?

Yes, there is another simple way to prove that a^3+1 is divisible by 7. We can use the fact that 7 is congruent to 1 modulo 3, which means that 7-1 is divisible by 3. Then, we can rewrite a^3+1 as a^3+(7-1), and using the property of congruence, we can simplify this to a^3+1 is congruent to a^3-1 (mod 3). Since a^3-1 is divisible by 3, a^3+1 must also be divisible by 3. And since 7 is also a factor of a^3+1, it is therefore divisible by 7.

## 4. Can this statement be generalized for any prime number?

Yes, this statement can be generalized for any prime number. Instead of using 7, we can use any prime number p and apply Fermat's Little Theorem. The proof would be similar, showing that a^p-a is divisible by p, and therefore a^3+1 would also be divisible by p.

## 5. Can this statement be applied to any integer, or only positive integers?

This statement can be applied to any integer, including negative integers. This is because the concept of divisibility remains the same for negative integers. For example, if we have -7 as the integer, it can be rewritten as (-1)^3+1, which is divisible by 7. Therefore, the statement still holds for negative integers as well.

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