Finding all solutions to quartic Diophantine equation

diophantine equationsnumber theory

Consider the Diophantine equation $6x^2 = y^2(2y-1)(y-1)$. I am interested in finding all solutions to this such that $y$ is a positive integer — or at the very least knowing whether there are infinitely many. Certainly there are some; the smallest being $(x,y) = (0,1)$, and then (less trivially) $(x,y) = (350,25)$. Generating more is possible.

This Diophantine equation arises in my research on certain determinants; I am not fluent enough in this area to see any immediate ways this question might be resolved. I had a look in Mordell, but could not see anything that could be shaped to give any direct answer (though perhaps I am wrong!).

Best Answer

Using the change of variables $ z = \frac{x}{y}$, we have $ 6z^2 = (2y-1)(y-1)$.
Completing the Square on RHS gives us $ 48z^2+1 = (4y-3)^2 $.
This is a Pell's equation with solutions $$ z = \frac{ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n } { 8 \sqrt{3} }, 4y-3 = \frac{1}{2} [ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n] .$$
Show that $y$ is an integer iff $n $ is even.
Hence, conclude that the solutions are .... (Just backtrack. I'm too lazy to paste the expressions.)
The first non-trivial solution is $ n = 2, 4y-3 = 97, z = 14, y = 25, x = 350$.
The next solution is $ n = 4, 4y-3 = 18817, z = 2716, y = 4705, x = 12778780$.

Previous version:

Using the change of variables $ z = \frac{x}{y}$, we have $ 6z^2 = (2y-1)(y-1)$.
Treating this as a quadratic in $y$, we have $y = \frac{1}{4} ( 3 \pm \sqrt{ 48z^2 + 1 } )$. We have a solution if the expression is a perfect square.
Finally, Pell's equation on $a^2 = 48z^2 + 1$ gives us $ z = \frac{ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n } { 8 \sqrt{3} }$ and $ a = \frac{1}{2} [ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n] $
It remains to backtrack (and check for positivity / integer-ness).

Related Solutions

Prove that the Pell’s Equation $x^2 −Dy^2 = 1$ always has a solution where $y$ is a multiple of $41$

Let $x_n+y_n\sqrt D=(x_1+y_1\sqrt D)^n$ the $n^{\text{th}}$ power of the primitive unit. Since there are only $41^2=1681$ possibilities for $(x_n,y_n)$ $\pmod{41}$ a duplicate must be encountered at some point: $x_n\equiv x_m\pmod{41}$ and $y_n\equiv y_m\pmod{41}$ for some $n>m\ge1$. Then $x_{n-m}=x_nx_m-Dy_ny_m\equiv x_n^2-Dy_n^2\equiv1\pmod{41}$ and $y_{n-m}=-x_ny_m+y_nx_m\equiv-x_ny_n+y_nx_n\equiv0\pmod{41}$.

EDIT: As an example let $D=3$ and the first solution to Pell's equation is $x_1+y_1\sqrt D=2+1\sqrt3$. Now let's make a table of values $\pmod{41}$: $$\begin{array}{r|r|r}n&x_n&y_n\\\hline 1&2&1\\ 2&7&4\\ 3&26&15\\ 4&15&15\\ 5&34&4\\ 6&39&1\\ 7&40&0\\ 8&39&40\\ 9&34&37\\ 10&15&26\\ 11&26&26\\ 12&7&37\\ 13&2&40\\ 14&1&0\\ 15&2&1\end{array}$$ For example $(2+1\sqrt3)^2=7+4\sqrt3$, $(2+1\sqrt3)^3=26+15\sqrt3$, and $(2+1\sqrt3)^4=97+56\sqrt3$ so $x_4=97\equiv15\pmod{41}$ and $y_4=56\equiv15\pmod{41}$, thus explaining the row $n=4$, $x_n\equiv15$, $y_n\equiv15$. The first duplicate was $x_{15}\equiv x_1\equiv2\pmod{41}$ and $y_{15}\equiv y_1\equiv1\pmod{41}$, so that tells us that $x_{15-1}=x_{14}\equiv1\pmod{41}$ and $y_{15-1}=y_{14}\equiv0\pmod{41}$. Perhaps a bit anticlimactic since we already found $2$ solutions on our way to generating the first duplicate. Indeed $x_{14}^2-3y_{14}^2=50843527^2-3\cdot29354524^2=1$ and $y_{14}=29354524=41\cdot715964$.

EDIT: Oh yeah, the last $2$ lines: since $(x_n+y_n\sqrt D)(x_n-y_n\sqrt D)=(x_1+y_1\sqrt D)^n(x_1-y_1\sqrt D)^n=(x_1^2-Dy_1^2)^n=(1)^n=1$ we see that $(x_n+y_n\sqrt D)^{-1}=(x_n-y_n\sqrt D)$ so $(x_{n-m}+y_{n-m}\sqrt D)=(x_n+y_n\sqrt D)(x_m-y_m\sqrt D)=(x_nx_m-Dy_ny_m)+(-x_ny_m+y_nx_m)\sqrt D$

EDIT My program that finds the fundamental solution to $x^2-Dy^2=1$ and the first power $n-m$ for which $x_{n-m}\equiv1\pmod{41}$ and $y_{n-m}\equiv0\pmod{41}$

program pell
   use ISO_FORTRAN_ENV
   implicit none
   integer(INT64) D
   integer(INT64) sqD, r, s, a, p0, p1, p, q0, q1, q, n
   integer(INT64) m
   write(*,'(a)') '  D         x_1                  y_1            n-m'
   do D = 1, 100
      sqD = int(sqrt(D+0.5D0),INT64)
      if(sqD**2==D) cycle
      r = 0
      s = 1
      p0 = 0
      p1 = 1
      q0 = 1
      q1 = 0
      do n = 1, 200
         a = (sqD+r)/s
         p = a*p1+p0
         p0 = p1
         p1 = p
         q = a*q1+q0
         q0 = q1
         q1 = q
         r = a*s-r
         s = (D-r**2)/s
         if(mod(n,2) == 0 .AND. s == 1) then
            write(*,'(i4,1x,i17,1x,i18)',advance='no') D,p,q
            p0 = mod(p,41)
            q0 = mod(q,41)
            p1 = 1
            q1 = 0
            do m = 1, 1000000
               p = p1*p0+D*q1*q0
               q = p1*q0+q1*p0
               p1 = mod(p,41)
               q1 = mod(q,41)
               if(p1 == 1 .AND. q1 ==0) then
                  write(*,'(1x,i4)') m
                  exit
               end if
            end do
            exit
         end if
      end do
   end do
end program pell

And its output:

  D         x_1                  y_1            n-m
   2                 3                  2    5
   3                 2                  1   14
   5                 9                  4   20
   6                 5                  2   42
   7                 8                  3   21
   8                 3                  1    5
  10                19                  6   20
  11                10                  3   42
  12                 7                  2    7
  13               649                180   14
  14                15                  4    7
  15                 4                  1   21
  17                33                  8   42
  18                17                  4    5
  19               170                 39   42
  20                 9                  2   20
  21                55                 12   40
  22               197                 42   42
  23                24                  5   10
  24                 5                  1   42
  26                51                 10   42
  27                26                  5   14
  28               127                 24   21
  29              9801               1820   14
  30                11                  2   42
  31              1520                273    5
  32                17                  3    5
  33                23                  4   40
  34                35                  6   21
  35                 6                  1   42
  37                73                 12   20
  38                37                  6   42
  39                25                  4   40
  40                19                  3   20
  41              2049                320   82
  42                13                  2   40
  43              3482                531   10
  44               199                 30   21
  45               161                 24   10
  46             24335               3588   20
  47                48                  7    7
  48                 7                  1    7
  50                99                 14    5
  51                50                  7   20
  52               649                 90   14
  53             66249               9100   14
  54               485                 66   14
  55                89                 12    7
  56                15                  2    7
  57               151                 20   40
  58             19603               2574   42
  59               530                 69   10
  60                31                  4   21
  61        1766319049          226153980    5
  62                63                  8   20
  63                 8                  1   21
  65               129                 16   42
  66                65                  8   10
  67             48842               5967   42
  68                33                  4   42
  69              7775                936   14
  70               251                 30   42
  71              3480                413   21
  72                17                  2    5
  73           2281249             267000   20
  74              3699                430   20
  75                26                  3   14
  76             57799               6630   21
  77               351                 40   40
  78                53                  6    8
  79                80                  9    7
  80                 9                  1   20
  82               163                 18   82
  83                82                  9    4
  84                55                  6   40
  85            285769              30996    2
  86             10405               1122   20
  87                28                  3   40
  88               197                 21   42
  89            500001              53000   42
  90                19                  2   20
  91              1574                165   40
  92              1151                120    5
  93             12151               1260    7
  94           2143295             221064    3
  95                39                  4    7
  96                49                  5   21
  97          62809633            6377352   42
  98                99                 10    5
  99                10                  1   42

Diophantine equation $\frac{x^4 + y^4}{z^4 + t^4} = u^2$

Here is a table with small solutions ($x \le y \le 7300$, ordered by $y$) for the equation $$x^4+y^4 = u^2(z^4+t^4);$$

it is divided into $3$ parts:

$u=1$;
$u=41$;
other values for $u$.

\begin{array}{|r|r|r|r|r|} \hline x & y & z & t & u \\ \hline 59 & 158 & 133 & 134 & 1 \\ 7 & 239 & 157 & 227 & 1 \\ 193 & 292 & 256 & 257 & 1 \\ 271 & 502 & 298 & 497 & 1 \\ 103 & 542 & 359 & 514 & 1 \\ 222 & 631 & 503 & 558 & 1 \\ 76 & 1203 & 653 & 1176 & 1 \\ 878 & 1381 & 997 & 1342 & 1 \\ 1324 & 2189 & 1784 & 1997 & 1 \\ 1042 & 2461 & 2026 & 2141 & 1 \\ 248 & 2797 & 2131 & 2524 & 1 \\ 1034 & 2949 & 1797 & 2854 & 1 \\ 1577 & 3190 & 2345 & 2986 & 1 \\ 1623 & 3494 & 2338 & 3351 & 1 \\ 661 & 3537 & 2767 & 3147 & 1 \\ 3364 & 4849 & 4288 & 4303 & 1 \\ 2694 & 4883 & 3966 & 4397 & 1 \\ 604 & 5053 & 1283 & 5048 & 1 \\ 2027 & 6140 & 4840 & 5461 & 1 \\ 274 & 6619 & 5093 & 5942 & 1 \\ 2707 & 6730 & 3070 & 6701 & 1 \\ 498 & 6761 & 5222 & 6057 & 1 \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ \hline 2 & 13 & 1 & 2 & 41 \\ 364 & 685 & 88 & 95 & 41 \\ 1223 & 2977 & 149 & 467 & 41 \\ 1417 & 5078 & 346 & 787 & 41 \\ 5146 & 5803 & 169 & 1022 & 41 \\ 8206 & 9633 & 894 & 1637 & 41 \\ ? & ? & ? & ? & ? \\ \hline 38 & 43 & 1 & 2 & 569 \\ 121 & 205 & 5 & 7 & 809 \\ 159 & 269 & 9 & 13 & 409 \\ 151 & 388 & 7 & 8 & 1889 \\ 9 & 437 & 1 & 3 & 21089 \\ 314 & 863 & 1 & 2 & 182209 \\ 859 & 1186 & 1 & 2 & 385241 \\ 10 & 1261 & 5 & 22 & 3281 \\ 1112 & 1323 & 7 & 12 & 14089 \\ 908 & 1559 & 7 & 8 & 31841 \\ 1041 & 2219 & 3 & 7 & 101201 \\ 2277 & 2521 & 1 & 3 & 905761 \\ 1166 & 2589 & 9 & 38 & 4729 \\ 2148 & 2723 & 12 & 23 & 15929 \\ 1383 & 2936 & 7 & 12 & 58049 \\ 1006 & 3177 & 2 & 3 & 1029961 \\ 911 & 3633 & 21 & 59 & 3769 \\ 475 & 3823 & 5 & 7 & 265721 \\ 1951 & 5777 & 79 & 103 & 2729 \\ 1897 & 5983 & 71 & 127 & 2129 \\ 2962 & 6741 & 22 & 57 & 14089 \\ 4969 & 6987 & 77 & 111 & 4001 \\ 4719 & 7187 & 1 & 3 & 6211649 \\ \cdots & \cdots & \cdots & \cdots & \cdots \end{array}

I'm curious why $u=41$ is relatively rich with solutions (after $u=1$). I have doubts if there are infinitely many solutions for $u=41$.
And, of course, I see no parametrization of solutions (to have a family of them).

Best Answer

Related Solutions

Prove that the Pell’s Equation $x^2 −Dy^2 = 1$ always has a solution where $y$ is a multiple of $41$

Diophantine equation $\frac{x^4 + y^4}{z^4 + t^4} = u^2$

Related Question