When is $8x^2-4$ a square number?

I asked an earlier question on when $32x+32$ is a square number (here) and I got a very clear answer.

Now I am looking to solve for which $x$ the equation $8x^2-4$ results in a square number. When I am trying to solve it I get: $y^2=4(2x^2-1)$ so if we put $y=2w$ then $2x^2-1 = w^2$ and therefore $x=\sqrt{\frac{w^2+1}{2}}$ which is correct.

Only now I need to check when $\frac{w^2-1}{2}$ is a square which goes on recursively. I am looking for Integer solutions for $x$ and $w$. How can I solve this?

Solution 1:

Given the equation $$ 2 n^2 - 1 = ( m - n )^2. $$

From this we can write $$ 2 m n = m^2 - n^2 + 1. $$

Let $$ x = m^2 - n^2, y = 2 m n, z = m^2 + n^2 \Rightarrow x^2 + y^2 = z^2. $$

But $$ y = x + 1, $$

so we obtain $$ x^2 + ( x + 1 )^2 = z^2, $$

which can be written as $$ 2 z^2 - 1 = ( 2 x + 1 )^2, $$

or $$ 2 \big( \underbrace{m^2 + n^2}_{n'} \big)^2 - 1 = \big( \underbrace{ 2 \big[ m^2 - n^2 \big] + 1 }_{m'-n'} \big)^2. $$

So starting with $$ \big( m, n \big) $$

we can "generate" the next pair $(m',n')$ using $$ \big( m', n' \big) = \big( 3 m^2 - n^2 + 1, m^2 + n^2 \big). $$

Note that $m=2$ and $n=1$ yields $$ 2 \cdot 1^2 - 1 = ( 2 - 1 )^2 $$

We can now "generate" the next pairs...

$$ \big( m', n' \big) = \big( 3 \cdot 2^2 - 1^2 + 1, 2^2 + 1^2 \big) = \big( 12 , 5 \big). $$

$$ \big( m'', n'' \big) = \big( 3 \cdot 12^2 - 5^2 + 1, 12^2 + 5^2 \big) = \big( 408 , 169 \big). $$

$$ \big( m''', n''' \big) = \big( 3 \cdot 408^2 - 169^2 + 1, 408^2 + 169^2 \big) = \big( 470832, 195025 \big). $$

The number we are looking for are the numbers $n$, the first given by $$ n = 1\\ n = 5\\ n = 169\\ n = 195025 $$

This are not all numbers, as the is a second method to generate pairs $(m',n')$.

I am still working on the second method.

$$ 2 x_n^2 - 1 = y_n^2, $$

where both $x_n$ and $y_n$ are integers, then $$ x_n = \frac{7 + 5 \sqrt{2}}{2 \sqrt{2}} \Big( 3 + 2 \sqrt{2} \Big)^{n-1} -\frac{7 - 5 \sqrt{2}}{2 \sqrt{2}} \Big( 3 - 2 \sqrt{2} \Big)^{n-1}. $$

The results are $$ \begin{array}{l|l} n & x_n\\ \hline 0 & 1\\ 1 & 5\\ 2 & 29\\ 3 & 169\\ 4 & 985\\ 5 & 5741\\ 6 & 33461\\ 7 & 195025\\ 8 & 1136689\\ 9 & 6625109\\ 10 & 38613965 \end{array} $$

(need to post how to derive it, but it is long)

Given $$ 2 x_n^2 - 1 = y_n^2. $$

Note that $$ \left( \begin{array}{c} x_{n+1} \\ y_{n+1} \end{array} \right) = \left( \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right) \left( \begin{array}{c} x_n \\ y_n \end{array} \right) $$

and $$ \left( \begin{array}{c} x_0 \\ y_0 \end{array} \right) = \left( \begin{array}{c} 1 \\ 1 \end{array} \right). $$

So $$ \left( \begin{array}{c} x_n \\ y_n \end{array} \right) = \left( \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right)^n \left( \begin{array}{c} 1 \\ 1 \end{array} \right). $$

Let $\chi$ be the trace of the matrix and $\Delta$ the determinant, then the eigenvalues are given by $\lambda_\pm = \chi/2 \pm \sqrt{\chi^2/4 - \Delta}$.

Then $$ \left( \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right)^n = \frac{\lambda_+^n - \lambda_-^n}{\lambda_+ - \lambda_-} \left( \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right) - \lambda_+ \lambda_- \frac{\lambda_+^{n-1} - \lambda_-^{n-1}}{\lambda_+ - \lambda_-} \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) $$

From this follows that $$ x_n = 5 \frac{\lambda_+^n - \lambda_-^n}{\lambda_+ - \lambda_-} - \lambda_+ \lambda_- \frac{\lambda_+^{n-1} - \lambda_-^{n-1}}{\lambda_+ - \lambda_-} $$

and $$ y_n = 7 \frac{\lambda_+^n - \lambda_-^n}{\lambda_+ - \lambda_-} - \lambda_+ \lambda_- \frac{\lambda_+^{n-1} - \lambda_-^{n-1}}{\lambda_+ - \lambda_-}. $$

Working this out we get $$ x_n = \frac{7 + 5 \sqrt{2}}{2 \sqrt{2}} \Big( 3 + 2 \sqrt{2} \Big)^{n-1} -\frac{7 - 5 \sqrt{2}}{2 \sqrt{2}} \Big( 3 - 2 \sqrt{2} \Big)^{n-1}. $$

(I have written this post fast, so forgive me for some typos :)

Solution 2:

As Daniel Fischer has said, you end up with $w^2 - 2x^2 = -1$, which is a negative Pell equation. I asked a question very similar earlier. The fundamental and minimal solution is $(w_1, x_1) = (1, 1)$. The rest of the solutions can be solved as convergents of the continued fraction of $\sqrt{2}$. The next solutions are described by the recurrence equations $w_{k+1} = 3w_k + 4x_k,\ x_{k+1} = 2w_k + 3x_k$. This resolves to $x_k = 6x_{k-1} - x_{k-2}$, with $x_0 = 0$. More information here.

Solution 3:

Continued Fraction Approach

Suppose $2q^2-1=p^2$. Then $$ \left(\sqrt2-\frac pq\right)\left(\sqrt2+\frac pq\right)=\frac1{q^2}\tag{1} $$ Thus, $$ \left|\sqrt2-\frac pq\right|\lt\frac1{2q^2}\tag{2} $$ The only rational approximations that are this good are Continued Fraction approximations. The continued fraction for $\sqrt2$ is $(1;2,2,2,2,\dots)$. Using the table below, where each column below the line is the sum of the the number above that column times previous column plus the column preceding that: $$ \begin{array}{c|c} &&1&2&2&2&2&2&2\\\hline 0&1&1&3&7&17&41&99&239\\ 1&0&1&2&5&12&29&70&169 \end{array}\tag{3} $$ the first several convergents for this continued fraction are $$ \color{#00A000}{\frac11},\frac32,\color{#00A000}{\frac75},\frac{17}{12},\color{#00A000}{\frac{41}{29}},\frac{99}{70},\color{#00A000}{\frac{239}{169}},\cdots\tag{4} $$ Under- and over-estimates alternate, Thus, the green entries are under-estimates.

Since the numerators and denominators follow the recurrence $$ (S^2-2S-1)a=0\tag{5} $$ where $S$ is the shift operator $Sa_n=a_n+1$, and $(x^2-2x-1)(x^2+2x-1)=x^4-6x^2+1$, the numerators and denominators must also satisfy $$ (S^4-6S^2+1)a=0\tag{6} $$ Therefore, the recurrence for every other numerator and denominator is $$ a_n=6a_{n-2}-a_{n-4}\tag{7} $$

Thus, if we start out with $a_1=1$ and $a_2=5$, then compute successive terms with $$ a_n=6a_{n-1}-a_{n-2}\tag{8} $$ we get all integers so that $8a_n^2-4=4(2a_n^2-1)$ is a square.

Notes
$(1)$: If $q\ne0$, this is equivalent to $2q^2-1=p^2$.
$(2)$: Since $p\ge q$, we actually have $\left|\sqrt2-\frac pq\right|\le\frac1{(1+\sqrt2)q^2}\lt\frac1{2q^2}$. The claim about rational approximations is Theorem $5.6$ from this paper.
$(3)$: This table was generated as described in the preceding paragraph. It is simply using the Wallis Algorithm, in particular Corollary $2.2$, from this paper.
$(4)$: This list simply collects the convergents from table $(3)$ and colors the under-estimates green.
$(5)$: This equation is an operator-based restatement of $a_n=2a_{n-1}+a_{n-2}$.
$(6)$: Any sequence that satisfies the constraint in $(5)$ also satisfies this equation.
$(7)$: This is a restatement of $(6)$, which is a relation among every other element of the sequence. This gives us a recurrence on the numerators and denominators of the under-estimates.
$(8)$: Describe the sequence of denominators for the under-estimates.