Is it true that the equation $27x^2+1=7^3y^2$ has infinitely many solutions in positive integers $x,y$ ?
Is it true that the equation $27x^2+1=7^3y^2$ has infinitely many solutions in positive integers $x,y$ ?
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From this site,
$27 x^2 - 343 y^2 +1 = 0$
by Dario Alejandro Alpern
X0 = 1 342879 (7 digits)
Y0 = 376766
If (x,y) is a solution, (-x,-y) is also a solution.
Xn+1 = P Xn + Q Yn
Yn+1 = R Xn + S Yn
P = 97 379496 466615 (14 digits)
Q = 347 082488 429404 (15 digits)
R = 27 321362 062956 (14 digits)
S = 97 379496 466615 (14 digits)
Here is some output from a program that anyone can write, it is Lagrange's method for indefinite forms. First, the cycle for $343s^2 - 27 t^2$ shows that it really does represent $1.$ This is in Dickson (1929) Introduction to the Theory of Numbers. I like the presentation in D. A. Buell, Binary Quadratic Forms. Program in C++, eventually I added in some features for oversize integers using GMP.
You Can Do This Yourself.
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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 343 0 -27
0 form 343 0 -27 delta -3
1 form -27 162 100
0 -1
1 -3
To Return
-3 1
-1 0
0 form -27 162 100 delta 1 ambiguous
1 form 100 38 -89 delta -1
2 form -89 140 49 delta 3
3 form 49 154 -68 delta -2
4 form -68 118 85 delta 1
5 form 85 52 -101 delta -1
6 form -101 150 36 delta 4
7 form 36 138 -125 delta -1
8 form -125 112 49 delta 3
9 form 49 182 -20 delta -9
10 form -20 178 67 delta 2
11 form 67 90 -108 delta -1
12 form -108 126 49 delta 3
13 form 49 168 -45 delta -4
14 form -45 192 1 delta 192
15 form 1 192 -45 delta -4 ambiguous
16 form -45 168 49 delta 3
17 form 49 126 -108 delta -1
18 form -108 90 67 delta 2
19 form 67 178 -20 delta -9
20 form -20 182 49 delta 3
21 form 49 112 -125 delta -1
22 form -125 138 36 delta 4
23 form 36 150 -101 delta -1
24 form -101 52 85 delta 1
25 form 85 118 -68 delta -2
26 form -68 154 49 delta 3
27 form 49 140 -89 delta -1
28 form -89 38 100 delta 1
29 form 100 162 -27 delta -6
30 form -27 162 100
form -27 x^2 + 162 x y 100 y^2
minimum was 1rep x = 212581 y = -376766 disc 37044 dSqrt 192.46817919 M_Ratio 50.81481
Automorph, written on right of Gram matrix:
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Let's see, as Tito pointed out, we also want to solve $u^2 - 21^3 v^2 = 1,$
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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell 9261
0 form 1 192 -45 delta -4
1 form -45 168 49 delta 3
2 form 49 126 -108 delta -1
3 form -108 90 67 delta 2
4 form 67 178 -20 delta -9
5 form -20 182 49 delta 3
6 form 49 112 -125 delta -1
7 form -125 138 36 delta 4
8 form 36 150 -101 delta -1
9 form -101 52 85 delta 1
10 form 85 118 -68 delta -2
11 form -68 154 49 delta 3
12 form 49 140 -89 delta -1
13 form -89 38 100 delta 1
14 form 100 162 -27 delta -6
15 form -27 162 100 delta 1
16 form 100 38 -89 delta -1
17 form -89 140 49 delta 3
18 form 49 154 -68 delta -2
19 form -68 118 85 delta 1
20 form 85 52 -101 delta -1
21 form -101 150 36 delta 4
22 form 36 138 -125 delta -1
23 form -125 112 49 delta 3
24 form 49 182 -20 delta -9
25 form -20 178 67 delta 2
26 form 67 90 -108 delta -1
27 form -108 126 49 delta 3
28 form 49 168 -45 delta -4
29 form -45 192 1 delta 192
30 form 1 192 -45
disc 37044
Automorph, written on right of Gram matrix:
236875798327 45535603438260
1011902298628 194522117134903
Pell automorph
97379496466615 9371227187593908
1011902298628 97379496466615
Pell unit
97379496466615^2 - 9261 * 1011902298628^2 = 1
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9261 3^3 * 7^3
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Name the matrix called "Pell Automorph" with the letter $A.$ Note that the first solution of $u^2 - 9261 v^2 = 1$ is just the left hand column of $A.$ There are infinitely many solutions, they can all be written as the left hand column of the matrix $A^n,$ for integer exponent $n \geq 1.$
I. Yes.
$$27x^2 + 1 = 7^3y^2\tag1$$
where,
$$x =\color{brown}{1342879} u^2 + 686\cdot\color{green}{376766} u v + 21^3\cdot\color{brown}{1342879} v^2$$
$$y =\color{green}{376766} u^2 + 54\cdot\color{brown}{1342879} u v + 21^3\cdot\color{green}{376766} v^2$$
and $u,v$ satisfy the Pell equation,
$$u^2-21^3v^2=1$$
II. In general:
If you have an initial solution $p,q$ to $mp^2-nq^2=d$ for any constant $d$, then an infinite more can be found by solving the Pell equation $u^2-mnv^2=\pm1$ and using the identity,
$$mx^2-ny^2 = d(u^2-mnv^2)^2\tag2$$
where,
$$x = pu^2+2nquv+mnpv^2$$
$$y = qu^2+2mpuv+mnqv^2$$
Thus, $(2)$ explains why we should set $u^2-mnv^2=\pm1$.
III. Even more generally:
If you have an initial solution $p,q$ to $ap^2+bpq+cq^2=d$, then an infinite more can be found by solving $u^2-Dv^2=\pm1$ with the familiar $D=b^2-4ac$ and using,
$$ax^2+bxy+cy^2 = d(u^2-Dv^2)^2\tag3$$
where,
$$x = p u^2 - 2(b p + 2c q)u v + D p v^2$$
$$y= q u^2 + 2(2a p + b q) u v + D q v^2$$
and $u,v$ solves $u^2-Dv^2=\pm1$, where $D=b^2-4ac$.