I'm looking for all solutions, (x,y,s,t) in the integers, for the two simultaneous equations... $$ 7x^2 - y^2 = 3s^2\\ 7y^2 - x^2 = 3t^2 $$ I have two solutions $(x,y,s,t) = (2,1,3,1)$ and $(751,422,1121,477)$.
I'm also interested in solving the more general cases... $$ Ax^2 + By^2 = Cs^2\\ Ay^2 + Bx^2 = Ct^2 $$ where $A + B = 2C$
Is there a heading I can search under for more info?
Thanks.
On
With regard to the general problem \begin{equation*} 2Ax^2+2By^2=(A+B)s^2 \hspace{2cm} 2Ay^2+2Bx^2=(A+B)t^2 \end{equation*} solutions do NOT exist for any combination of $A$ and $B$.
For example, for $A=5, B=-1$, there are no solutions since the quadric $5x^2-y^2=2s^2$ is not locally soluble at the prime $p=5$.
Allan Macleod
On
Above equation shown below:
$\begin{split} Ax^2 + By^2 = Cs^2\\B x^2 + A y^2 = Ct^2\end{split}$
The above simultaneous equation has parametric solution given by Mr. Seiji Tomita & is shown below.
$x$ = $4(5625m^{14}$-1075$m^{12}$$n^2$+9509$m^{10}$$n^4$+2553$m^8n^6$-101$m^6$$n^8$-137$m^4$$n^{10}$+ 7$m^2$$n^{12}$+3$n^{14}$)$m^2$
$y$ = ($n^2$+3$m^2$)(-$n^{14}$-$m^2$$n^{12}$+27$m^4$$n^{10}$+427$m^6$$n^8$+2173$m^8$$n^6$-8291$m^{10}$$n^4$+ 16425$m^{12}$$n^2$+5625$m^{14}$)
$s$ = (-$n^2$+5$m^2$)(-$n^{14}$-$m^2$$n^{12}$+27$m^4$$n^{10}$-597$m^6$$n^8$-4995$m^8$$n^6$-7267$m^{10}$$n^4$-9175$m^{12}$$n^2$+5625$m^{14}$)
$t$ = 4($n^{14}$+5$m^2$$n^{12}$-83$m^4$$n^{10}$-271$m^6$$n^8$-269$m^8$$n^6$- 6049$m^{10}$$n^4$+6175$m^{12}$$n^2$+16875$m^{14}$)mn
Where $(A,B,C)=[(16m^2),(n^2-9m^2),(7m^2+n^2)]$
For $(m,n)=(2,1)$ we have
$(x,y,s,t)=[(1570223344),(1969901167),(870604529),(2363928872)]$
$(A,B,C)=(64,-35,29)$
Seiji Tomita's article can be viewed on his web site
"Computational number theory" article # 306 & the link is given below:
On
The general case of the OP, \begin{align} Ax^2+By^2 &=Cs^2\\ Ay^2+Bx^2 &=Ct^2 \end{align} with, \begin{align} A &=(p+q)^2-2q^2\\ B &=(p-q)^2-2q^2\\ C &= \,p^2-q^2 \end{align} where the OP used $(p,q) = (2,1)$ so $(A,B,C) = (7,-1,3)$ leads to an elliptic curve with rank at least $1$ for any $(p,q)$ with $|p| \ne |q|$ and $pq \ne 0$. This gives an infinite number of possible parametric solutions using $(p,q)$. One such is
\begin{equation*} x=q(3p^8-4p^6q^2+14p^4q^4+4p^2q^6-q^8) \end{equation*} \begin{equation*} y=p(p^8-4p^6q^2-14p^4q^4+4p^2q^6-3q^8) \end{equation*}
The derivation is a standard (if boring) computation, using a symbolic algebra package.
On
Above equation shown below:
$Ax^2 + By^2 = Cs^2\\ Ay^2 + Bx^2 = Ct^2$
"OP" want's parametric solution for variable's $(x,y,s,t)$. Allen Macleod has kindly provided solution only for variable's $(x,y)$. By extension, the parametric solution given by Allen Macleod for variables $(s,t)$ would be:
$s=(p-q)(p^8+16p^5q^3+14p^4q^4+16p^3q^5+q^8)$
$t=(p+q)(p^8-16p^5q^3+14p^4q^4-16p^3q^5+q^8)$
For $(p,q)=(2,1)$ we get back solution given by "OP" as
$(A,B,C)=(7,-1,3)$
$(x,y,s,t)=(751,422,1121,477)$
There are an infinite number of solutions, but they get large quite quickly. Solving this system is a standard application of elliptic curves.
The quadric $7x^2-y^2=3s^2$ has the simple solution $x=1, y=2, s=1$, which allows us to find the parametric solution $x=3k^2-6k+7, y=2(3k^2-7)$.
Substituting into $7y^2-x^2=3t^2$ gives the quartic \begin{equation*} t^2=81k^4+12k^3-418k^2+28k+441 \end{equation*} which has an obvious rational point when $k=0$, and so is birationally equivalent to an elliptic curve.
Standard methods find this curve to be \begin{equation*} v^2=u^3-97u^2+2352u \end{equation*} with \begin{equation*} k=\frac{6v-u}{3(9u-448)} \end{equation*}
The elliptic curve has $7$ finite torsion points at $(0,0)$, $(48,0)$, $(49,0)$, $(42,\pm 42)$ and $(56, \pm 56)$. It also has rank $1$ with generator $(21,126)$. This generator gives $k=-35/37$, and hence the second solution quoted.
Doubling the generator gives the following solution \begin{equation*} x=124344271, \, y=56190422, \, s=187147999, \,t=47046243 \end{equation*}
Further computation can give more solutions.
Allan Macleod