$8x^2-7=y^2$
I tried this:
$8(x^2-1)=y^2-1$
$8(x^2-1)=(y-1)(y+1)$
Let $x^2-1 = a*b$
I made cases:
$y-1 = 8a , y+1 = 1b$
$y-1 = 4a , y+1 = 2b$
$y-1 = 2a , y+1 = 4b$
$y-1 = 1a , y+1 = 8b$
but wasn't able to continue further.
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7$\begingroup$ $y^2-2X^2=-7$ is a generalized Pell equation $\endgroup$J. W. Tanner– J. W. Tanner2024-10-30 03:04:40 +00:00Commented Oct 30, 2024 at 3:04
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1$\begingroup$ solutions to that include $(X_n,y_n)$, where $(1+2\sqrt2)(3+2\sqrt2)^n=y_n+X_n\sqrt2$ $\endgroup$J. W. Tanner– J. W. Tanner2024-10-30 03:12:30 +00:00Commented Oct 30, 2024 at 3:12
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1$\begingroup$ other solutions are $(X_n,y_n)$ where $(5+4\sqrt2)(3+2\sqrt2)^n=y_n+X_n\sqrt2$; cf. wikipedia $\endgroup$J. W. Tanner– J. W. Tanner2024-10-30 03:24:58 +00:00Commented Oct 30, 2024 at 3:24
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3$\begingroup$ also see OEIS $\endgroup$J. W. Tanner– J. W. Tanner2024-10-30 03:38:49 +00:00Commented Oct 30, 2024 at 3:38
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1$\begingroup$ Note that $(3+2\sqrt2)^n=y_n+X_n\sqrt2$ are solutions to $y^2-2X^2=1$ $\endgroup$J. W. Tanner– J. W. Tanner2024-10-30 04:18:15 +00:00Commented Oct 30, 2024 at 4:18
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2 Answers
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as 7 is prime this is not too bad.
Rather than many interwoven sequences of generalized Fibonacci type, here there are just two subsequences of $x_n, y_n$ with $x_n^2 - 8 y_n^2 = -7.$ They obey the same recurrence,
$$ x_{n+4} = 6 x_{n+2} - x_n , $$
$$ y_{n+4} = 6 y_{n+2} - y_n . $$
We could say that one subsequence is odd $n$ and the other even $n.$ $$ $$ $$\left( \begin{array}{ccccccccccccc} n &-1 & 0 & 1 & 2 & 3 & 4 &5 & 6 & 7 & 8 & 9 &10 & 11 \\ \hline x_n & -5 & -1 & 1 & 5 & 11 & 31 &65 & 181 & 379 & 1055 & 2209 & 6149 & 12875 \\ y_n & 2 & 1 & 1 & 2 & 4 & 11 & 23 & 64 & 134 & 373 & 781 & 2174 & 4552 \\ \end{array} \right) $$
Put slightly differently, any solution $(x,y)$ there is a new solution at $(3x+8y, x+3y)$ which gives larger values if $x,y,> 0.$
In the other direction, if you have a solution $(x,y)$ with large positive $x,y,$ a new solution with saller positive $(x,y)$ by taking $(3x-8y, -x+3y).$ Going backwards in this way, we must reach a solution with $(x,y > 0$ but at least one of $(3x-8y, -x+3y)$ negative or zero. The two "seed" solutions are thus $(1,1)$ which backs up to $(-5,2),$ also $(5,2)$ which backs up to $(-1,1).$
For such "seed" solutions, the system $x,y > 0$, with $3x < 8y,$ and $x^2 - 8 y^2 = -7$ comes out to the inequality $$ x < \sqrt {56} \; \approx \; 7.48 $$
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$\begingroup$ Your answer is very nice, but for clarity I'm pointing out that you switched the role of $x$ and $y$ from OP $\endgroup$J. W. Tanner– J. W. Tanner2024-10-30 21:12:15 +00:00Commented Oct 30, 2024 at 21:12
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$\begingroup$ Also, what about the solutions where both $x$ and $y$ are negative? $\endgroup$J. W. Tanner– J. W. Tanner2024-10-30 21:18:48 +00:00Commented Oct 30, 2024 at 21:18
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The key to solving this equation is to recognize it as a generalized Pell equation: $y^2-2X^2=-7.$
$($I write $X=2x.)\qquad$
Solutions to the regular Pell equation $y^2-2X^2=1$ are given by $(X,y)=(X_n,y_n),$
where $y_n+X_n\sqrt2=(3+2\sqrt2)^n, n=0,1,2,3,...\qquad$ Solutions to $y^2-2X^2=-7$
are given by $(X,y)=(X_n,y_n)$, where $y_n+X_n\sqrt2=(1+2\sqrt2)(3+2\sqrt2)^n, n=0,1,2,3,...$
and by $(X,y)=(X_m,y_m), $ where $y_m+X_m\sqrt2=(5+4\sqrt2)(3+2\sqrt2)^m, m=0,1,2,3,...$
The solutions $y$ are sequence A077446 in The On-Line Encyclopedia of Integer Sequences.
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$\begingroup$ I gave solutions for positive integers; obviously the signs of $X$ and/or $y$ could be changed and they would still be integer solutions $\endgroup$J. W. Tanner– J. W. Tanner2024-10-30 21:20:19 +00:00Commented Oct 30, 2024 at 21:20