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Given that, for integers $p$ and $q$, $$\begin{align} A &= p^2-q^2+2pq \\[2pt] B &= p^2-q^2-2pq \\[2pt] C &= p^2-q^2 \end{align}$$

it can readily be shown that the two conic sections

$$\begin{align} Ax^2 + By^2 &= Cs^2 \\[2pt] Ay^2 + Bx^2 &= Ct^2 \end{align}$$

have a trivial parametric solution $$(x, y, s, t ) = ( p, q, p + q, p - q )$$

My question is

How does one generate nontrivial parametric solutions?

PWJVE

Blue
  • 75,673

3 Answers3

0

Above simultaneous equation's shown below:

$\begin{align} Ax^2 + By^2 &= Cs^2 \\[2pt] Ay^2 + Bx^2 &= Ct^2 \end{align}$

Conditions are:

$\begin{align} A &= p^2-q^2+2pq \\[2pt] B &= p^2-q^2-2pq \\[2pt] C &= p^2-q^2 \end{align}$

"OP" suggested solution is:

$(x, y, s, t ) = ( p, q, p + q, p - q )$

Above solution has non trivial solutions also:

For example, $(p,q)=(7,4)$ we get:

$(x, y, s, t ) = ( 7,4,11,3) )$

And $(a,b,c)=(89,-23,33)$

Sam
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Above equation shown below:

$\begin{align} Ax^2 + By^2 &= Cs^2 \\[2pt] Ay^2 + Bx^2 &= Ct^2 \end{align}$

Mr. Seiji Tomita has given solution to the above simultaneous equation :

Where, $A = 16m^2, B = n^2-9m^2, C=7m^2+n^2$

$x = 4m^2$

$y = 3m^2+n^2$

$s = 5m^2-n^2$

$t = 4mn$

For $(m,n)=(3,2)$ the solution is:

$(x,y,s,t)=( 36,31,41,24)$ and

$(A, B, C)=(144,-77, 67 )$

His solution can be viewed on his website, article #306 & the link is given below:

    http://www.maroon.dti.ne.jp/fermat/eindex.html
Sam
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Question asked by 'OP' has solution shown below:

$(x,y,s,t)=(751,422,1121,477)$

$(A,B,C)$ =$[(7),(-1),(3)]$

The solution can be at the below mentioned link:

https://math.stackexchange.com/questions/3108936
Sam
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