A System of Simultaneous Pell Equations

Question

Are $0$ and $\pm1$ the only integer solutions for which both $\sqrt{24n^2+1}$ and $\sqrt{48n^2+1}$ are simultaneously integers ?

---

Whilst pondering upon the Biblical concept of the Jubilee year, I couldn't help but notice that a time period of $7^2=49$ years, apart from approximating half of a decimal century of $10^2=100$ years, also lies conspicuously close to a third of a duodecimal “century” of $12^2=144$ years. In other words, I was left with solving the system of Diophantine equations $$\dfrac{x^2}3+1~=~y^2~=~\dfrac{z^2}2-1,$$ which, after a bit of basic number theory, boiled down to the system of Pell equations described above. Then, a Mathematica search of depth up to $10^4$ into the $($ periodic $)$ continued fraction expansion of $~\lim\limits_{n\to\infty}\sqrt{\dfrac{48n^2+1}{24n^2+1}}~=~\sqrt2~$ failed to reveal any other solutions, save for the ones mentioned earlier, implying that any other possible values of n possess at least $3,828$ digits.

---

• A $2007$ paper by Mihai Cipu and Maurice Mignotte shows that there might be at most one such positive solution, for $n>1.$

• In the same year, the $34^{th}$ volume of the Annales Mathematicae et Informaticae, published by the Eszterhazy Karoly University of Applied Sciences, contains a relevant paper by Laszlo Szalay on the resolution of simultaneous Pell equations.

Answer 1

The problem is related to the Cannonball problem in Wikipedia. In more detail, the book Unsolved Problems in Number Theory Second Edition by Richard Guy in section D3 starting on page 223:

The case "square pyramid = square" is Lucas's problem. Is $\, x=24,\ y=70 \,$ the only nontrivial solution of the diophantine equation $$ y^2 = x(x+1)(2x+1)/6? $$ This was solved affirmatively by Watson, using elliptic functions, and by Ljunggren, using a Pell equation in a quadratic field. Mordell asked if there was an elementary proof, and affirmative answers have been given by Ma, by Xu & Cao, by Anglin and by Pinter.

The same equation in disguise is to ask if $\, (48,140) \,$ is the unique nontrivial solution to the case "square = tetrahedron", since the previous equation may be written $$ (2y)^2 = 2x(2x+1)(2x+2)/6, $$ though, as Peter Montgomery notes, this doesn't eliminate the possibility of an odd square. A more modern treatment is to put $\, 12x = X-6,\ 72y = Y \,$ and note that $\, Y^2 = X^3 - 36X \,$ is curve $576H2$ in John Cremona's tables. The point $\, (12,36) \,$ (which gives an odd square) serves as a generator. There's an infinity of rational solutions, but the only nontrivial integer solution to the original equation is given by the point $\, (294,5040). \,$

The modern treatment that Richard Guy is writing about here uses the elliptic curve with Cremona label $\texttt{"576h2"}$ or LMFDB label 576.c3 which has two $2$-torsion point generators $(0,0)$ and $(6,0)$. The point $(0,0)$ added to $(-3,9)$ is $(12,36)$ where the latter two points are infinite order generators. The other integral points are $\ (-6,0), (-2,8), (18,72), (294,5040). \ $ This is directly related to the congruent number problem. In particular, finding rational right triangles with area $6$. Read about this in, for example, Karl Rubin on Right triangles and elliptic curves. Another method is given in my work at Weierstrass Elliptic Function Polynomials.

In brief, I construct four polynomial sequences $\, (w_n,x_n,y_n,z_n) \,$ such that $\, (x_n/x_0)^2 - (y_n/y_0)^2 = (x_1^2-y_1^2) (w_n/w_1)^2, \,$ $\, (z_n/z_0)^2 - (y_n/y_0)^2 = (z_1^2-y_1^2) (w_n/w_1)^2, \,$ $\, (z_n/z_0)^2 - (z_n/z_0)^2 = (z_1^2-x_1^2) (w_n/w_1)^2 \,$ for all integer $\, n.$ They satisfy recursions such as $\, w_{n+1}w_{n-1}x_0^2 = w_n^2 x_1^2 - x_n^2 w_1^2, \,$ and $\, x_{n+1}x_{n-1}x_0^2 = x_n^2 x_1^2 - (x_1^2-y_1^2)(x_1^2-z_1^2)(w_n/w_1)^2 x_0^4. $

To get solutions of $\, j^2 = 1 + 24m^2, \, k^2 = 1 + 48m^2 \,$ we let $\, x_0 = y_0 = z_0 = w_1 = x_1 = 1 \,$ and $\, y_1 = \sqrt{2}, \, z_1 = \sqrt{3}. \,$ This is almost exactly what I used in my answer to MSE question 2556314. Let $\, t_n := w_n/\sqrt{24}. \,$ The solutions are given by $\, j = y_{2n}/x_{2n}, \, k = z_{2n}/x_{2n}, \, m = t_{2n}/x_{2n}. \,$ For $\, n=0, \,$ we have a trivial solution $\, (1)^2 = 1 + 24(0)^2, \, (1)^2 = 1 + 48(0)^2. \,$ For $\, n=1, \,$ we have a nontrivial solution $\, (-5)^2 = 1 + 24(-1)^2, \, (-7)^2 = 1 + 48(-1)^2. \,$ For $\, n=2, \,$ we have $\, (-1201/1151)^2 = 1+24(70/1151)^2, \, (-1249/1151)^2 = 1+48(70/1151)^2. \, $ For all $\, n>1 \,$ the solutions are non-integer rationals.

Now the cannonball connection is that solutions of $\, y^2 \!=\! x(x\!+\!1)(2x\!+\!1)/6 \,$ come from letting $\, x\!=\!24m^2, \, x\!+\!1\!=\!j^2, \, 2x\!+\!1\!=\!k^2, \, y \!=\! 2mjk \,$ from solutions of $\, j^2 \!=\! 1 \!+\! 24m^2, \, k^2 \!=\! 1 \!+\! 48m^2 .\,$ Thus, if only one nontrivial integer solution of this equation exists, then the cannonball problem has only one nontrivial solution and this has been proven by several authors and methods.

Answer 2

Added: from Cipu and Mignotte, the first system expected to have two positive solutions, one piece $\color{red}{x^2 - 24 z^2 = 1},$ uses $490^2 - 1 = 240099.$ $$ 5^2 - 24 \cdot 1^2 = 1 \; , \; \; \; 490^2 - 240099 \cdot 1^2 = 1 $$ $$ 4801^2 - 24 \cdot 980^2 = 1 \; , \; \; \; 480199^2 - 240099 \cdot 980^2 = 1 $$

The first system expected to have two positive solutions, one piece $\color{red}{x^2 - 48 z^2 = 1},$ uses $1358^2 - 1 = 1844163.$ $$ 7^2 - 48 \cdot 1^2 = 1 \; , \; \; \; 1358^2 - 1844163 \cdot 1^2 = 1 $$ $$ 18817^2 - 48 \cdot 2716^2 = 1 \; , \; \; \; 3688327^2 -1844163 \cdot 2716^2 = 1 $$ $$ $$ $$ $$

I ran the thing for my $j \leq 1148961$ with $k \leq 1000000,$ or $$ x \approx y \approx \color{magenta}{10^{1143895}} $$ no matches

Probably the most accurate way to search, just integer arithmetic, with a little cleverness to keep track of comparable subscripts $j,k$ so as not to waste time comparing $x_j$ and $y_k$ when the ratio cannot be close to $1.$

Second try: as there are just two sequences, in order to avoid saving long stretches, we could just run both sequences up a few terms. Then, if the current $x_j < y_k,$ take more terms $x_j$ until we have exceeded the stationary $y_k.$ Next, increase the $k$ until $y_k$ is bigger. The computer will not need any array, just three x terms and three y terms.

You are asking whether these two sequences have any elements larger than $1$ in common:

$$ 0, 1, 10, 99, 980, 9701, 96030, ... $$ $$ x_{j+2} = 10 x_{j+1} - x_j $$

$$ 0, 1, 14, 195, 2716, 37829, 526890, ... $$ $$ y_{k+2} = 14 y_{k+1} - y_k $$

================================

Thu May  3 10:02:52 PDT 2018
  j  3   x   99
  k  3   y   195
  j  4   x   980
  k  4   y   2716
  j  5   x   9701
  k  5   y   37829
  j  6   x   96030
  k  6   y   526890
  j  7   x   950599
  k  7   y   7338631
  j  8   x   9409960
  k  8   y   102213944
  j  10   x   922080050
  k  9   y   1423656585
  j  11   x   9127651499
  k  10   y   19828978246
  j  12   x   90354434940
  k  11   y   276182038859
  j  13   x   894416697901
  k  12   y   3846719565780
  j  14   x   8853812544070
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  j  16   x   867583274883920
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  j  18   x   85014307126080090
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  k  17   y   2016350381665827089
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  k  18   y   28084137899285228670
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  k  62   y   6023449585620939154423742663309618525796051217702134904088053332603314
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  k  64   y   1168518170076240447542240590036153662610679659069193935414528342682094016
  j  74   x   4814550834280912967057671826280394096544946962380416512719699586967264370
  k  65   y   16275358551091519023684463665313332470754052676233936805066352769262552129
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  k  66   y   226686501545205025884040250724350500927946057808205921335514410426993635790
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  k  74   y   321060162067922620610547530906016238786499979345421401197804493668540503920914490374
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  k  76   y   62284016447913134193552370964684997450481766154908407321244338625654162098671091177876
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  j  99   x   37352216856094034277028263209462688201614577106839709332676419855814424786086081807069785021122499
  k  87   y   238354116298331613822159001920283541166990434467447765638220463369048230244939081531970000479645655
  j  100   x   369748828398900702004712076028419368034192131612969278087333593446012224595953848156189910924824500
  k  88   y   3319844572575886772975928730522917142859399782093862444005360462799267241581842643184746870769304936
Thu May  3 10:02:52 PDT 2018

==================================

int main()
{
  system("date");

  mpz_class x, xx, xxx;
  mpz_class y,yy,yyy;
  x = 10; xx = 1; 

  y = 14; yy = 1;


  mpz_class j,k;
  j = 2;
  k = 2;

  int goon = 1;
  while ( goon && j < 100)
  {
     while (goon && x < y)
     {
       xxx = 10 * x - xx ;
       xx = x;
       x = xxx;
       ++j;
     }
     cout << "  j  "  << j << "   x   "  << x << endl;
     if( x == y) 
      {
         cout <<  "   WOW  "  << endl;
         goon = 0 ;
      }
    while( goon && y < x)
    {
       yyy = 14 * y - yy;
       yy = y;
       y = yyy;
       ++k ;
    }
     cout << "  k  "  << k << "   y   "  << y << endl;
      if( x == y) 
      {
         cout <<  "   WOW  "  << endl;
         goon = 0 ;
      }
  }

  system("date");
  return 0;
}

==================================

Answer 3

A wild guess:

Write the equivalent system $$x^2= 24 n^2 + 1 \\ y^2 = 48 n^2 + 1$$ so $$x^2 + 24 n^2 = y^2$$ This equation has some ( families ) of solutions that are parametrized by some pairs of integers $(a,b)$. Express $x$, $y$ in terms of $a$, $b$ (quadratically) and plug in into the first equation. We get an equation in $a$, $b$ of degree $4$, a Thue equation. There are some effective bounds for the solutions of such an equation, although they may be too large to check up tot them. Perhas this is the method in the last paper you quoted.

Added: From the first equation we conclude $(x^2, 24 n^2)=1$, and so $(x^2, z^2)=1$. From the third equation we get $$(y-x)(y+x)= 24 n^2$$. $y$ and $x$ are both odd. We conclude that $(y-x, y+x)=2$. Therefore, we have several possibilities:

$$\begin{array}{c:c:c:c:c} \frac{y-x}{2} & 6 u^2& 3 u^2 & 2 u^2& u^2 \\ \frac{y+x}{2} & v^2 & 2v^2 & 3 v^2 & 6 v^2 \\ n & u v& u v& u v & u v \end{array} $$

Answer 4

Here a general solution of this problem using Fermat's method of infinite descent (the simplest version of it currently very sophisticated). From $$24n^2+1=x^2\\48n^2+1=y^2$$ we have, since $x$ and $y$ are odd and $x\lt y$

$$(x+2h)^2=y^2\Rightarrow h^2+hx-6n^2=0\Rightarrow h= \frac{-x\pm\sqrt{x^2+24n^2}}{2}=$$ Because of $24n^2=x^2-1$ the radicand becomes $2x^2-1$ and must be equal to $t^2$ so we have $$2x^2-1=t^2\iff1+t^2=2x^2$$ From identity $$(r^2-s^2+2rs)^2+(r^2-s^2-2rs)^2=2(r^2+s^2)^2$$ comes the general solution of the equation $$X^2+Y^2=2Z^2..........(*) $$ it follows, because $(1,t,x)$ is a solution of $(*)$ $$\begin{cases}r^2-s^2+2rs=t \\r^2-s^2-2rs=1\iff2r^2=1+(r+s)^2 \\r^2+s^2=x\Rightarrow r\text{ and }s\lt x \end{cases}$$ This shows another solution $(1,r+s,r)$ of $(*)$ with $r\text{ and }s\lt x$ so we have $$\begin{cases}r_1^2-s_1^2+2r_1s_1=r+s \\r_1^2-s_1^2-2r_1s_1=1\iff2r_1^2=1+(r_1+s_1)^2 \\r_1^2+s_1^2=r\Rightarrow r_1\text{ and } s_1\lt r\end{cases} $$ Again another solution of $(*)$ with $r_1\text{ and }s_1\lt r$ and the procedure can be iterated how many times you want.

By descent the proof is done.

Answer 5

Not an answer yet, only a comment. I was not able to arrive at an answer but this work might give someone else more experienced than me an idea how to proceed

Step a: Let's denote the expression with the multiplicator $24$ as the diophantine equation $$ 24x^2 +1 = t^2$$ and we look at the set of $x_h$ with natural index $h \ge 0$ such that we get a set of perfect squares of $t_h$
That set is $X=\{0,1,10,99,980,... x_h,...\}$ and we can find

• the recursion-formula $x_{h+2}=10 x_{h+1} - 1 x_h $ beginning with $x_0=0,x_1=1$ or

• the $\sinh()$ formula using the following constants: $s_q=\sqrt{5^2-1}$ , $q = 5 + s_q$, $\quad l_q= \ln q$ and then $$x_h=f(h)=\sinh(l_q \cdot h)/s_q $$

Step b: We can do the same with the expression with $48$ as multiplicator: $$ 48y^2 +1 = u^2$$ and we look at the set of $y_i$ with natural index $i \ge 0$ such that we get a set of perfect squares of $u_i$
That set is $Y=\{0,1,14,195,2716,... y_i,...\}$ and we can find

• the recursion-formula $y_{i+2}=14 \cdot y_{i+1} - 1 y_i $ beginning with $y_0=0,y_1=1$ or

• the $\sinh()$ formula using the following constants: $s_r=\sqrt{7^2-1}$ , $r = 7 + s_r$, $\quad l_r= \ln r$ and then $$y_i=g(i)=\sinh(l_r \cdot i)/s_r $$

---

Step c: Now I see two options, but I can't finish any of them.

Opt c1: Using the $\sinh()$ formula one can formulate $$ get\_h(i)\quad = f°^{-1}(g(i)) \qquad= \sinh°^{-1}( \sinh(l_r \cdot i)/s_r \cdot s_q)/l_q $$ This gives the fractional index into a generalized set $X$ (when generalized to the reals) for any $y_i$ at integer index $i$ in $Y$.
Computing the generalized $h$ from the integer $i$ such that $x_h = y_i$ by $h=get\_h(i)$ gives some heuristical data:
(update: added more information in following table)

   i   h(generalized)  i'=(h+d_24)/c_24      err1=i'-i           h'=i*c_24-d_24     err2 = h - h'
  -----------------------------------------------------------------------------------------------------

    0            0.0   0.13158112005           0.131581120053  -0.151181644749           0.151181644749
    1  1.00000000000   1.00193228121         0.00193228120785   0.997779883231         0.00222011676906
    2  2.14675300150   2.00001008762       0.0000100876239708    2.14674141121       0.0000115902918512
    3  3.29570299894   3.00000005200    0.0000000520028797854    3.29570293919    0.0000000597493082176
    4  4.44466446748   4.00000000027  0.000000000268063295398    4.44466446717  0.000000000307994413475
    5  5.59362599515   5.00000000000        1.38180628232E-12    5.59362599515        1.58764225751E-12
    6  6.74258752313   6.00000000000        7.12290205824E-15    6.74258752313        8.18394043250E-15
    7  7.89154905111   7.00000000000        3.67169655979E-17    7.89154905111        4.21863808961E-17
    8  9.04051057909   8.00000000000        1.89267738302E-19    9.04051057909        2.17461349797E-19
    9  10.1894721071   9.00000000000        9.75632822017E-22    10.1894721071        1.12096457793E-21
   10  11.3384336351   10.0000000000        5.02916879513E-24    11.3384336351        5.77832146333E-24
   11  12.4873951630   11.0000000000        2.59242393236E-26    12.4873951630        2.97859536250E-26
   12  13.6363566910   12.0000000000        1.33633650388E-28    13.6363566910        1.53539923139E-28
   13  14.7853182190   13.0000000000        6.88851552903E-31    14.7853182190        7.91463932775E-31
   14  15.9342797470   14.0000000000        3.55087555088E-33    15.9342797470        4.07981939861E-33
   15  17.0832412750   15.0000000000        1.83039685760E-35    17.0832412750        2.10305557032E-35
   16  18.2322028029   16.0000000000        9.43528605354E-38    18.2322028029        1.08407806810E-37
   17  19.3811643309   17.0000000000        4.86367874500E-40    19.3811643309        5.58817976246E-40
   18  20.5301258589   18.0000000000        2.50711751613E-42    20.5301258589        2.88058157216E-42
   19  21.6790873869   19.0000000000        1.29236295595E-44    21.6790873869        1.48487531658E-44
   20  22.8280489149   20.0000000000        6.66184173327E-47    22.8280489149        7.65419985702E-47
   21  23.9770104428   21.0000000000        3.43403028343E-49    23.9770104428        3.94556868158E-49
   22  25.1259719708   22.0000000000        1.77016573789E-51    25.1259719708        2.03385233098E-51
   23  26.2749334988   23.0000000000        9.12480811460E-54    26.2749334988        1.04840534739E-53
   24  27.4238950268   24.0000000000        4.70363431774E-56    27.4238950268        5.40429487278E-56
   25  28.5728565548   25.0000000000        2.42461819659E-58    28.5728565548        2.78579302792E-58
   26  29.7218180827   26.0000000000        1.24983640354E-60    29.7218180827        1.43601394393E-60
   27  30.8707796107   27.0000000000        6.44262687544E-63    30.8707796107        7.40233041901E-63
   28  32.0197411387   28.0000000000        3.32102993149E-65    32.0197411387        3.81573562456E-65
   29  33.1687026667   29.0000000000        1.71191658606E-67    33.1687026667        1.96692629650E-67
   30  34.3176641947   30.0000000000        8.82454677642E-70    34.3176641947        1.01390647480E-69

Here, the constant $c_{24}$ is $c_{24}=f°^{-1}(\sqrt 2)$ and $d_{24}=f°^{-1}(\sqrt 3 / 24)$.
We see in column 3 (i'), that a simple linear transformation of the generalized $h$ by the (irrational) constants $c_{24}$ and $d_{24}$ is enough to reproduce integer $i$ with a quickly vanishing error.
In reverse, the values $h'$ taken by the linear function $h'(i)=i \cdot c_{24}+d_{24}$ approximates $h(i)$ very quickly with vanishing error. Since $c_{24},d_{24}$ are likely transcendent values it should be impossible to equal an integer at any index $i$.

With this I tried to show, that the function $get\_h(i)$ is non-integer for all integer $i>0$ - but of course, such a conclusion needs now some more formal arguing using arguments of from theory of diophantine approximation.

Opt c2: We want, that at some $h>0$ and $i>0$ we find (at least) one equality $x_h = y_i$. Here we observe, that the primefactorization in $x_h$ and $y_i$ is cyclic with the indexes and each possible primefactors in $x_h=y_i$ has its specific cycle-length. The cycle-lengthes of the primefactors in $x_h$ correspond now with primefactors in $h$ and that of $y_i$ with primefactors in $i$.
Here is now the task to show that the cycle-lengthes for $x_h$ and $y_i$ are different enough so that we do not find any combination of $h$ and $i$ such that $x_h=y_i$ by their canonical primefactorization.
This allows to exclude very large $x_h$ and $y_i$ with little effort, but again I do not have a conclusive method to show that this has, so to say, an "infinite ascent".

Note, this comment is reflecting a related question of mine and by this a very similar question having only different parameters and which I've tried to answer the same way as written here.

Answer 6

Let $n$ be a nonnegative integer greater than $1$ such that $24n^2+1$ and $48n^2+1$ are both perfect squares. Then \begin{eqnarray} 24n^2+1&=&a^2,\\ 48n^2+1&=&b^2, \end{eqnarray} for positive integers $a$ and $b$. Clearly $n$ is coprime to $a$ and $b$, and we have $$a^2+24n^2=b^2.$$ This shows that $(n,a,b)$ is an integral point on a conic defined over $\Bbb{Q}$. The integral points on this conic have an integral parametrization by binary quadratic forms; in this case $(n,a,b)$ is of one of the following two forms: \begin{eqnarray} a&=&6s^2-t^2,\qquad&\text{ }&\qquad a&=&3s^2-2t^2,\\ n&=&st,\qquad&\text{ or }&\qquad n&=&st,\\ b&=&6s^2+t^2,\qquad&\text{ }&\qquad a&=&3s^2+2t^2. \end{eqnarray} Plugging these two parametrizations back into the first equation of our system yields $$ 1=36s^4-36s^2t^2+t^4 \qquad\text{ or }\qquad 1=9s^4-36s^2t^2+4t^4, $$ which are two Thue equations, for which there exist effective methods to find all integral solutions. PARI/GP tells me that the nonnegative integral solutions to the first equation are $(s,t)=(0,1)$ and $(s,t)=(1,1)$, corresponding to $n=0$ and $n=1$. The second equation has no nonnegative integral solutions.

This shows that $n=0$ and $n=\pm1$ are the only integral solutions to your problem.