# Solve for n to prove that the given equation has a solution.

• MHB
• Chris L T521
In summary, to solve for n in a given equation, you need to isolate it using algebraic operations and then plug in different values to see if they satisfy the equation. Any value for n can be used to prove the equation has a solution, but some values may be more helpful than others. If no value of n satisfies the equation, then the equation has no solution. An equation can have one, two, or infinite solutions depending on its complexity and variables. Proving that an equation has a solution may not always be necessary, but it can aid in understanding the relationship between variables and solving related problems.
Chris L T521
Gold Member
MHB
I would like to thank those of you that participated in the first POTW on MHB. Now it's time for round two! (Bigsmile)This week's problem was proposed by caffeinemachine.Problem: Let each of the numbers $a_1, a_2, \ldots , a_n$ equal $1$ or $-1$. If we're given that $a_1 a_2 a_3 a_4 + a_2 a_3 a_4 a_5 + \ldots + a_n a_1 a_2 a_3 =0$, prove that $4|n$.There were no hints provided for this problem.Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-(POTW)-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!EDIT: Since there was some confusion as to what this was asking, I included a clarification in the spoiler.

Here are some more terms in the sum: $a_1a_2a_3a_4 + a_2a_3a_4a_5 + a_3a_4a_5a_6+a_4a_5a_6a_7+\cdots+a_{n-2}a_{n-1}a_na_1 + a_{n-1}a_na_1a_2+a_na_1a_2a_3$.

You may now be asking what happens in the case of $n=4$. Well, if you follow this pattern, you are cycling through the indices 4 at a time, and once you get to $n$, you "start over again." So, if $n=4$, the indices of the terms in the sum will be a cyclic permutation of $(1234)$ [starting with $(1234)$ and ending with $(4123)$].

Last edited:
Congratulations to the following members for their correct solutions:

1) Sudharaka

Solution:

[sp]Claim: $n$ is even.
Proof: Since each of $a_i's$ is either $1$ or $-1$ each of the summands in the LHS of the given equation is also either $1$ or $-1$. Since the RHS is $0$ the number of $1's$ appearing in the LHS is equal to the number of the $-1's$ appearing in the LHS. Thus $2|n$.

Claim: $4|n$.
Proof: From the above claim we can write $n=2k$.
Consider $(a_1a_2a_3a_4)(a_2a_3a_4a_5) \ldots (a_na_1a_2a_3)=(a_1a_2 \ldots a_n)^4$ .
In the above RHS is definitely $1$ while LHS is $(-1)^k (1)^k$.
Thus $(-1)^k(1)^k=1 \Rightarrow (-1)^k=1 \Rightarrow 2|k.$ Hence $4|n$.[/sp]

## 1. How do I solve for n to prove that the given equation has a solution?

To solve for n, you need to isolate it on one side of the equation by using algebraic operations such as addition, subtraction, multiplication, and division. Once you have n on its own, you can plug in different values to see if they satisfy the equation.

## 2. Can I use any value for n to prove that the given equation has a solution?

Yes, you can use any value for n as long as it satisfies the equation. However, some values may be more useful in proving that the equation has a solution than others.

## 3. What if I can't find a value for n that satisfies the equation?

If you have exhausted all possible values for n and none of them satisfy the equation, then the equation does not have a solution. This means that there is no value of n that makes the equation true.

## 4. How many solutions can an equation have?

An equation can have one, two, or infinite solutions. It depends on the complexity of the equation and the number of variables involved.

## 5. Is it necessary to prove that an equation has a solution?

No, it is not always necessary to prove that an equation has a solution. Some equations are already known to have solutions, while others may not have a solution at all. However, proving that an equation has a solution can help in understanding the relationship between the variables and can also be useful in solving other related problems.

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