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Solve (Find all the solutions, if there are, or prove there are not) the system of 2 Pell or Pell-like simultanous equations over the positive integers :

$2b^{2}= a^{2} +1 = 3k^{2} + 2 $ with the 3 variables (a,b,k).

Note that the 2 equations are : $a^{2} -2b^{2} = -1$ and $a^{2} -3k^{2} = 1$ The third one: $ 2b^{2} -3k^{2} =2$ does not need to be solved if the other two are.

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there is $(a,b,k) = (7,5,4)$, and I would be surprised if there was any other solution. –  mercio Jan 19 '13 at 21:13
    
Yes mercio (7,5,4) is a solution. Try to find more or prove there are not any more. –  user55514 Jan 19 '13 at 21:39

3 Answers 3

László Szalay, On the resolution of simultaneous Pell equations, Annales Mathematicae et Informaticae 34 (2007) pp. 77–87, is available here. I don't know whether the precise pair of equations you give is discussed in the paper, but it does give you the tools and references you need to approach this kind of problem.

EDIT: Having looked a bit more closely at the paper, I believe a solution to your problem is in fact given in the paper and attributed to Riele, reference 24.

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Here is the solution to the simultanous $$ \begin{align}3x^2−y^2=2\tag{1}\\8x^2−z^2=7\tag{2}\end{align} $$ $$ x_{n+1}=2x_n+y_n\quad\text{and}\quad y_{n+1}=3x_n+2y_n $$ for $(1)$ and $$ x_{n+1}=3x_n+z_n\quad\text{and}\quad z_{n+1}=8x_n+3z_n $$ for $(2)$. And we get: $(1)\Rightarrow(x,y)=(1,1);(3,5);(11,19);(41,71);(153,265);(571,989)$ and $(2)\Rightarrow(x,z)=(−1,1);(2,5);(11,31);(64,181);(373,1055);(2174,6149)$. The solutions coincide in the same $n^{\text{th}}$ position, the third position: for the only solution $(x,y,z) = (11,19,31)$.

Clearly, x grows faster, after its values coincide, permanently in (2) than in (1), and thus we prove this is the only non trivial solution. The case (2) has another family of solutions : $(1,1);(4,11);(23,65);(134,379)$ which obviously still grow faster, beginning with a higher value of one of its initial terms and will never give a solution.

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Please use LaTeX to make the math readable. MathJax is the LaTeX interface supported here. –  robjohn Jan 26 '13 at 12:41
    
This is not an answer as it does not prove that there might not be further solutions. –  robjohn Jan 26 '13 at 13:09
    
This proves clearly the unicity of the non trivial solution. And i seldom use TeX cause i am a militant for a simplification of the "notation" and the "naming" of mathematics. I do prefer the quite correct and exact and simpler ASCII. How maths is written (drawn; in fact) does not alter its truth. –  user55514 Jan 27 '13 at 9:54
    
Although your examples indicate that the $x_n$ from $(2)$ grow faster than those from $(1)$, this is not a proof of that fact. However, even after that fact is established, why is it not possible that $x_{n_1}$ from $(1)$ might equal $x_{n_2}$ from $(2)$ for two different indices $n_1$ and $n_2$? –  robjohn Jan 27 '13 at 10:06
    
Although the truth of the math is the unchanged no matter how it is written, it is part of the purpose of this site to allow these truths to be understood by those who wish to understand them. In this function, making the math easier to read makes it easier to understand. –  robjohn Jan 27 '13 at 10:08

$a^2 - 3*k^2 = 1$ (1) , general solution:

$a(n+1) = 2a(n) + 3k(n)$, $k(n+1) = a(n) + 2k(n)$. The first 6 solutions, without counting the trivial solution: (1,0) are: $(a,k) = (2,1); (7,4); (26,15); (97,56); (362,209); (1351,780);\dots$

$a^2 - 2*b^2 = -1$ (2), general solution:

$a(n+1) = 3a(n) + 4b(n)$, $b(n+1) = 2a(n) + 3b(n)$. The first 6 solutions are: $(a,b) = (1,1); (7,5); (41,29); (239,169); (1393,985); (8119,5741);\dots$

Once we match the same a value in both equations: a=7 asociated to k=4 and b=5; we do not need to look any more for solutions, since both a and b grow faster in one equation than a and k in the other ---> the only solution non trivial is (a,b,k) = (7,5,4). The problem would be if a(n) grew faster and b(n) slower in one equation than in the other, or the contrary. In that case we would have to match another coincident value of "a" among the infinite values in both equations. But no problem here.

The general solution of the third equation $2b^{2} -3k^{2} = 2$ is $b(n+1) = 5b(n)+ 6k(n) ; k(n+1) = 4b(n) + 5k(n)$ with solutions: $(1,0);(5,4);(49,40);(485,396);(4801,3920), ...$

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2  
This is not a valid argument, as for all you know the 794th $a$-value from the first equation equals the 538th $a$-value from the second equation, and that would give you another solution to the pair of equations. Please look at the paper to which I linked in my answer to see what one has to do to solve this kind of system of equations. Also, if you are going to make regular use of this website, it would be good if you would learn something about formatting mathematics here. –  Gerry Myerson Jan 20 '13 at 23:07
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Please try to take a more civil tone on this website. Not only this not an answer, but your entire comment amounts to repeating your argument more forcefully in the hope that it somehow gains substance (which, by the way, it didn't). –  Erick Wong Jan 21 '13 at 6:44
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I understand what your claim is. Now consider the third solution $(x,y,z) = (97,1567,56)$. I hope you'll take a close look at what Gerry Myerson told you, with a more open mind this time. –  Erick Wong Jan 21 '13 at 18:35
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@Erick, how you were able to resist writing GOTCHA!, I'll never know. Well done. –  Gerry Myerson Jan 22 '13 at 11:55
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You can keep living in your fantasy world, or you can read the Szalay paper and actually learn some mathematics. Your choice. –  Gerry Myerson Jan 24 '13 at 0:10

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