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How to solve Simultaneous equation in Mathematica:

x^2 - (a^2 - 1)y^2 = 1 and y^2 - pz^2 = 1 where 0< a < 16 and 0 < y < 100 and 0 < p < 50 and 0 < z < 100.

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    $\begingroup$ What have you tried? $\endgroup$ – Daniel Lichtblau Jun 1 '18 at 15:46
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    $\begingroup$ It's very discouraging to see questions that show little effort. If you expect to inspire anybody to volunteer their time to look at your problem, then start by doing your part. Read the Mathematica documentation. Get the Informed badge in the site by taking and understanding the tour. Please write your equations in Wolfram Mathematica language (Also in $\LaTeX$ if that helps). To add new information edit your question, use the comments section just for comments. $\endgroup$ – rhermans Jun 1 '18 at 19:00
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(Referring to the answer by @chuy in "Make Reduce produce nicer output").

Regarding a and p as parameters and x, y, z as variables.

red = Reduce[{x^2 - (a^2 - 1) y^2 == 1 , y^2 - p z^2 == 1, 0 < a < 16,
 0 < y < 100 , 0 < p < 50 , 0 < z < 100}, {x, y, z}, Reals];

TraditionalForm[
   red //. Or -> 
   Composition[(Column[#, Right, Background -> {{White, LightGray}}, 
   Frame -> All] &), List]]

enter image description here

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Read the documentations for Solve and Assuming. And understand the difference between Set(=) and Equal(==).

Assuming[
 {0 < a < 16,
  0 < y < 100,
  0 < p < 50,
  0 < z < 100},
 Solve[{
   x^2 - (a^2 - 1) y^2 == 1,
     y^2 - p z^2 == 1
   }]
 ]

$\left\{\\ \left\{p\to \frac{y^2-1}{z^2},x\to -\sqrt{a^2 y^2-y^2+1}\right\},\\ \left\{p\to \frac{y^2-1}{z^2},x\to \sqrt{a^2 y^2-y^2+1}\right\},\\ \{x\to -a,y\to -1,z\to 0\},\\ \{x\to a,y\to -1,z\to 0\},\\ \{x\to -a,y\to 1,z\to 0\},\\ \{x\to a,y\to 1,z\to 0\}\\ \right\}$

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    $\begingroup$ To @rhermans. I think, since the OP gives conditons for p and z , the pz is simply a typo. You should regard it as p*z . $\endgroup$ – Akku14 Jun 1 '18 at 19:37
  • $\begingroup$ @Akku14 thanks, I think you are right. It's corrected now. $\endgroup$ – rhermans Jun 1 '18 at 19:42

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