# Solving 6n^2-18n+16 with Pell's/Diophantus Equation

In summary, the equation 6n^2-18n+16 = m^2 can be put into the form 3/2 (2n-3)^2 + 5/2 = m^2. From this, we can solve for n and m, with the potential use of Pell's equation. The solution for n is given as 3/2 +- 1/2 \sqrt{2/3 ( m^2 - 5/2 )}, and the solution for m is given as +- \sqrt{3/2 (2n-3)^2 + 5/2}. To find integer solutions, we can set 2/3 ( m^2 - 5/2 )

## Homework Statement

Solve the equation 6n^2-18n+16 = m^2

the solution should be from pell's equation or Diophantus equation:
a^2 x^2 + c = y^2

## The Attempt at a Solution

I put my equation in the form: 3/2 (2n-3)^2 + 5/2 = m^2

Thanks

For which variable should you solve? I assume for n, then from 3/2 (2n-3)^2 + 5/2 = m^2 I get (2n-3)^2 = 2/3 ( m^2 - 5/2 ) and then 2n-3 = +- \sqrt{2/3 ( m^2 - 5/2 )}, so n = 3/2 +- 1/2 \sqrt{2/3 ( m^2 - 5/2 )}. If you should solve for m, m = +- \sqrt{3/2 (2n-3)^2 + 5/2}.
But since that's trivial, I assume there are further conditions on m and n. Should they be rational? Or integers?
An integer solution is, for example, n = 0 and m = +-4, or n = 1 and m = +-2.

OK, thanks for help

but when n = 3/2 +- 1/2 \sqrt{2/3 ( m^2 - 5/2 )}
how can i find the integer values for n?

(I assume you solve it without using pell equation)

I saw it, but the problem i think thay always have the factor of X^2 =1
and in my example dont

what if i change the form of my equation to:

2/5 m^2 - 3/5 (2n-3)^2 = 1

this is close to Pell equation form x^2 - P*y^2 = 1

Yes, I see. I'm sorry, don't know more.

From n = 3/2 +- 1/2 \sqrt{2/3 ( m^2 - 5/2 )}, you see that 2/3 ( m^2 - 5/2 ) must be a perfect odd square for n to be integer, so you can equally solve 2/3 ( m^2 - 5/2 ) = (2k+1)^2, with k = 0,1,2,... This can be reexpressed as m = +- \sqrt{6 k^2 + 6 k + 4}, so you can in principle find all solutions by calculating the above for all k and checking if it gives you an integer m ...
For k < 200, you have luck for k = 0 with m = 2, k = 1 with m = 4, k = 6 with m = 16, k = 15 with m = 38, k = 64 with m = 158, k = 153 with m = 376.
In principle, this gives you all solutions, but I don't see any general formula. For me, the only thing I can see is that 2k < m < 3k for large k.

## 1. What is Pell's/Diophantus Equation?

Pell's/Diophantus equation is a type of Diophantine equation, which is an equation that involves integer solutions. It is written in the form ax^2 + by^2 = c, where a, b, and c are integers and x and y are unknown variables.

## 2. How is Pell's/Diophantus Equation different from other types of equations?

Pell's/Diophantus equation is different because it specifically looks for integer solutions, rather than any real number solutions. This makes it a more challenging problem to solve, and it often requires the use of advanced techniques and mathematical tools.

## 3. How can Pell's/Diophantus Equation be used to solve 6n^2-18n+16?

First, we can rewrite the equation as 6n^2-18n+16 = 0, which is in the form ax^2 + by^2 = c. Then, we can use the techniques of Pell's/Diophantus equation to find integer solutions for n.

## 4. What are the steps to solving 6n^2-18n+16 with Pell's/Diophantus Equation?

The steps to solving this equation would involve finding the continued fraction representation of the square root of 6, using that to generate solutions for the equation, and then testing those solutions to find the one that satisfies the original equation. This process may need to be repeated multiple times to find all possible solutions.

## 5. What are some real-world applications of Pell's/Diophantus Equation?

Pell's/Diophantus equation has been used in cryptography, specifically in the RSA algorithm for secure communication. It has also been used in the study of prime numbers and in the construction of Pythagorean triples, which are sets of three positive integers that satisfy the Pythagorean theorem.

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