Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Suppose I want x and y to be rationals

Solve[ x^2 + y^2 == 1, {x, y}, Rationals]

I am told:

Solve::svars: Equations may not give solutions for all "solve" variables. 

and given a "solution"

{{y -> ConditionalExpression[-Sqrt[ 1 - x^2], (x | y) ∈ Rationals && -1 <= x <= 1]},
 {y -> ConditionalExpression[ Sqrt[ 1 - x^2], (x | y) ∈ Rationals && -1 <= x <= 1]}}

Try Reduce

Reduce[ x^2 + y^2 == 1, {x, y}, Rationals]

Mathematica returns

(x | y) ∈ Rationals && -1 <= x <= 1 && (y == -Sqrt[1 - x^2] || y == Sqrt[1 - x^2])

Not much more help. Next try to see what Mathematica can do make a list of solutions:

FindInstance[ x^2 + y^2 == 1, {x, y}, Rationals, 10]

I am told that:

FindInstance::fwsol: Warning: FindInstance found only 2 instance(s), but it was not able
to prove 10 instances do not exist.

and given the two solutions found

{{x -> -1, y -> 0}, {x -> 1, y -> 0}}

Mathematica did not help too much here - but does show me that at least some solutions exist and that there may be more.

  • Is this all I am going to get out of Mathematica in this case?
  • Is this an answer that a mathematican wants to know and Mathematica has done its work?
  • If I better understood Mathematica, could I use its other functions on the solution from Solve or Reduce to help me learn more about the solutions? (or do I just have to get my hands dirty and do the work myself - in this example it is pretty easy to find the general solution by hand)
  • And maybe more generally, which Mathematica functions are able to do something more with this type of output from Solve or Reduce? (e.g. can I plot it?)
share|improve this question
    
You already know the procedure for parametrically generating Pythagorean triples, I take it? –  J. M. Aug 2 '12 at 16:34
    
A naive way. Union@(Abs @ {x/y, z/w} /. FindInstance[(x/y)^2 + (z/w)^2 == 1 && 1 < x < 100 && x != y != z != w != 0, {x, y, z, w}, Integers, 10]) –  belisarius Aug 2 '12 at 17:10
1  
BTW, Mathematica is not a mathematician. It may help you, but it is not a replacement for your math knowledge. –  belisarius Aug 2 '12 at 19:06
add comment

1 Answer

Direct solutions using Solve

Since there are infinitely many solutions to the problem at hand we prefer to choose a bound on the solution space, i.e. we solve (x/y)^2 + (z/w)^2 == 1 in Integers and select only solutions satisfying x/y <= z/w and 0 < x <= y <= 50 and z <= w <= 50. It takes a bit but we know we have found all solutions satisying given criteria.

Union[{#1/#2, #3/#4} & @@@ Solve[(x/y)^2 + (z/w)^2 == 1 && 0 < x <= y <= 50 &&
                                  x/y <= z/w && z <= w <= 50, {x, y, z, w},
                                                                  Integers][[All, All, 2]]]
{{9/41, 40/41}, {7/25, 24/25}, {12/37, 35/37}, {5/13, 12/13}, {8/17, 15/17},
 {3/5, 4/5}, {20/29, 21/29}}

Now we can check if these are solutions :

And @@ (#1^2 + #2^2 == 1 & @@@ %)
 True

Well, indeed.

Edit 1

Solutions generated by Table of Pythagorean triples

Some more efficient way of generating rational solutions of the equation x^2 + y^2 == 1 is based on Pythagorean triples. We can write a table of numbers of the form e.g. $(\frac{x^2-y^2}{x^2+y^2},\frac{2 x y}{x^2+y^2})$ and select only distinct ones :

pts = Union @ Flatten[Table[{(x^2 - y^2)/(x^2 + y^2), (2 x y)/(x^2 + y^2)},
                            {y, 20}, {x, 20}], 1];
Length @ pts
255    
ParametricPlot[ {Cos[u], Sin[u]}, {u, 0, Pi/2}, PlotStyle -> Thick, 
                                   Epilog -> { Red, PointSize[0.006], Point @ pts}]

enter image description here

Solutions generated by Fibbonacci numbers

We could find only special solutions making use of observations by Dujella that Fibbonacci numbers $F_{n}$ generate distinct Pythagorean triples $(F_{n} F_{n+3},2F_{n+1}F_{n+2},F_{n+1}^2+F_{n+2}^2)$ although not exhaustively. Thus proceeding this way we can find arbitrary many distinct solutions quite effectively. We find the first 100 such solutions :

ptsF = 
  Table[{( Fibonacci[n] Fibonacci[n + 3])/( Fibonacci[n + 1]^2 + Fibonacci[n + 2]^2),
         ( 2Fibonacci[n + 1] Fibonacci[n + 2])/( Fibonacci[n + 1]^2 + Fibonacci[n + 2]^2)},
         {n, 100}];

And @@ (#1^2 + #2^2 == 1 & @@@ ptsF)
True

Let's write some of them :

ptsF[[;; 20]]

enter image description here

ParametricPlot[{Cos[u], Sin[u]}, {u, 0, Pi/2}, Epilog -> {Red, Point @ ptsF}]

enter image description here

Edit 2

Since there could be some kind of misunderstanding I add another way to visualize the solution set for various numbers of solutions constructed with tables of Pythagorean triples.

Clear[pt]
pt[k_Integer /; OddQ[k] && k > 0] := 
  Union @ Flatten[ 
            Table[{(x^2 - y^2)/(x^2 + y^2), (2 x y)/(x^2 + y^2)}, 
                                       {y, -k, k, 2}, {x, -k, k} ], 1]

Now we check what is distribution of solution points in various cases, e.g.

GraphicsRow[ Table[ 
                    ParametricPlot[ {Cos[u], Sin[u]}, {u, 0, 2 Pi},
                                                      PlotStyle -> Thin, 
                                                      Epilog -> { Red, Point @ p} ],
                   {p, {pt[9], pt[17], pt[55]}}]]

enter image description here

share|improve this answer
1  
Also Union@({(x/y), (z/w)} /. {ToRules[ Reduce[(x/y)^2 + (z/w)^2 == 1 && 0 < x <= y <= 50 && x/y <= z/w && z <= w <= 50, {x, y, z, w}, Integers]]}) –  belisarius Aug 2 '12 at 19:41
    
Thank you Belisarius for your guidance. The andser is helpful for solving this problem and I had not thought of this particular way of proceeding. –  Ron Burns Aug 2 '12 at 20:13
    
I want to add that I had used this particular example to get a solution with Domain restrictions. My question, while rather fuzzy to say the least, was what can I do with those kinds of solutions (with Domain or similar restictions) as input to other Mathematica Functions? –  Ron Burns Aug 2 '12 at 20:26
    
@RonBurns Consider registering your account, then you'll have more opportunities to benefit from using this site. Could you ask more precisely ? As I wrote there were infinitely many solutions, so to choose only specific ones you have to add conditions to Solve or Reduce. In this case there is no way to find all possible solutions. –  Artes Aug 2 '12 at 20:30
    
@Artes, "In this case there is no way to find all possible solutions" - that's why I suggested to the OP that he might want to use the rational parametrization for generating Pythagorean triples; the suggestion was not heeded, though. To reiterate belisarius: "Mathematica is not a mathematician. It may help you, but it is not a replacement for your math knowledge." –  J. M. Aug 3 '12 at 2:32
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.