Efficiently Solve Equations like a^n + b^n = c^n with These Tips

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In summary, the conversation discusses an equation of the form a^n + b^n = c^n and the attempt to solve for n. It is noted that Fermat's last theorem can be used as an upper bound for n and that 2 is a possible solution. The conversation also considers the possibility of n not being an integer and discusses the use of Binet's formula for the Fibonacci sequence. Ultimately, the goal is to find a way to solve the equation without using a computer.
  • #1
medwatt
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Hello.
I have encountered an equation of the form a^n + b^n = c^n. I know this looks like Fermat's equation. How can I solve for n. It doesn't matter if n is not an integer.
Thanks
 
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  • #2
Ok the equation I'm trying to solve is :

[itex]\left[\sqrt{2+\sqrt{3}}\right]^{n}[/itex]+ [itex]\left[\sqrt{2-\sqrt{3}}\right]^{n}[/itex] = [itex]2^{n}[/itex]
[itex]^{}[/itex]

I have arrived to this point:

[itex]\left[\sqrt{2+\sqrt{3}}\right]^{2n}[/itex] + [itex]\left[2*\sqrt{2+\sqrt{3}}]\right]^{n}[/itex] - 1 =0

I think the problem is easy but there is a point I'm missing.

I know that 2 is the answer from simple observation and with the knowledge that with Fermat's type of equations an integer solution can't be more than 2.
 
Last edited:
  • #3
So you want to solve

[itex]\left[\frac{\sqrt{2+\sqrt{3}}}{2}\right]^{n}+ \left[\frac{\sqrt{2-\sqrt{3}}}{2}\right]^{n}= 1[/itex]

and you know that n=2 is a solution, and I assume you know that an is decreasing for 0<a<1, so that there can't be more solutions.
 
  • #4
okay id try n=0, then n=1 then n=2... and see which ones or one are true. basically you're trying to find n.

also i think you are on the right track using fermat's last theorem (proved by Andrew Wiles in 1995) as an upper bound for n
 
  • #5
I need to show that n=2. The answer 2 in this problem was obvious. What would be the general solution to such equations if n is not an integer (i.2 we cannot guess it by simple observation).
 
  • #6
medwatt said:
Ok the equation I'm trying to solve is :

[itex]\left[\sqrt{2+\sqrt{3}}\right]^{n}[/itex]+ [itex]\left[\sqrt{2-\sqrt{3}}\right]^{n}[/itex] = [itex]2^{n}[/itex]
[itex]^{}[/itex]

I have arrived to this point:

[itex]\left[\sqrt{2+\sqrt{3}}\right]^{2n}[/itex] + [itex]\left[2*\sqrt{2+\sqrt{3}}]\right]^{n}[/itex] - 1 =0

I think the problem is easy but there is a point I'm missing.

I know that 2 is the answer from simple observation and with the knowledge that with Fermat's type of equations an integer solution can't be more than 2.

Do you need an exact analytic answer or will a numerical approximation suffice?
 
  • #7
Does n have to be an integer?
 
  • #8
It doesn't matter if n is an integer or not. I just want a way to solve the equation because suppose guess the answer was not so easy, how can one solve it without of course a computer.
 
  • #9
It resembles a Binet formula.
 
  • #10
Yes I think it looks like Binet's formula for Fibonacci sequence.
 

FAQ: Efficiently Solve Equations like a^n + b^n = c^n with These Tips

How do I know which method to use to solve an equation?

The method used to solve an equation depends on the type of equation. For linear equations, you can use methods like substitution or elimination. For quadratic equations, you can use the quadratic formula or factoring. It is important to understand the characteristics of different types of equations in order to choose the right method.

Why do we need to solve equations in math?

Solving equations is an essential skill in math as it helps us find the values of unknown variables. This is important in problem-solving and modeling real-life situations, such as calculating the cost of a purchase with tax or determining the height of a building using the Pythagorean theorem.

What is the order of operations in solving equations?

The order of operations in solving equations follows the acronym PEMDAS, which stands for Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right). This means that operations within parentheses are solved first, followed by exponents, then multiplication and division, and finally addition and subtraction.

How do I check if my solution to an equation is correct?

To check if your solution is correct, simply substitute the value you found for the variable back into the original equation. If the equation holds true, then your solution is correct. For example, if the equation is 2x + 3 = 7 and you found that x = 2, you would substitute 2 back into the equation and get 2(2) + 3 = 7, which is true.

Can equations have more than one solution?

Yes, equations can have more than one solution. This is called infinite solutions. For example, the equation x + 2 = x + 2 has infinite solutions because any value for x would make the equation true. On the other hand, equations can also have no solution, which means there is no value for the variable that would make the equation true.

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