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Just trying to figure out a way to generate triples for $a^2+b^2=5c^2$. The wiki article shows how it is done for $a^2+b^2=c^2$ but I am not sure how to extrapolate.

Gerry Myerson
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KaliMa
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3 Answers3

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This is one of those CW answers. Country and Western.

I did Gerry's recipe and I quite like how it works. Educational, you might say. I took the slope $t = \frac{q}{r}$ and starting rational point $(2,1).$ The other point works out to be $$ x = \frac{2 t^2 - 2 t - 2}{t^2 + 1}, \; \; \; y = \frac{- t^2 - 4 t + 1}{t^2 + 1}, $$ so multiply everything by $r^2$ to arrive at

$$ x = \frac{2 q^2 - 2qr - 2r^2}{q^2 + r^2}, \; \; \; y = \frac{- q^2 - 4 qr + r^2}{q^2 + r^2}. $$ So far $x^2 + y^2 = 5.$ Multiply through by $q^2 + r^2$ to get

$$ a = 2 q^2 - 2 q r - 2 r^2 $$

$$ b = -q^2 - 4 q r + r^2 $$

$$ c = q^2 + r^2 $$ $$ a^2 + b^2 = 5 q^4 + 10 r^2 q^2 + 5 r^4 $$ and $$ c^2 = q^4 + 2 r^2 q^2 + r^4 $$ and $$ a^2 + b^2 = 5 c^2 $$

Will Jagy
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Consider the circle $$x^2+y^2=5$$ Find a rational point on it (that shouldn't be too hard). Then imagine a line with slope $t$ through that point. It hits the circle at another rational point. So you get a family of rational points, parametrized by $t$. Rational points on the circle are integer points on $a^2+b^2=5c^2$.

Gerry Myerson
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  • I pick point (1,2) because 1^2+2^2=1+4=5. So i have a line with slope t going through it. What do I do with that? I can twist this line anywhere I want and have it go through any point on the other side of the circle. Slope 0 it hits (-1,2), etc – KaliMa Dec 03 '12 at 04:33
  • For each $t$, you get a point. For each rational $t$, you get a rational point. Then you clear denominators to get a triple. – Gerry Myerson Dec 03 '12 at 04:35
  • I am trying this but I am sorry, I don't see how any of this helps with the question. I am trying to create triplets like you see in http://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple – KaliMa Dec 03 '12 at 04:47
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    @KaliMa In the same link that you give, if you scroll down a little to the section Geometry of Euclid's Formula you'll find precisely what Gerry is trying to explain to you in the case of the usual Pythagorean Triples. – Adrián Barquero Dec 03 '12 at 05:00
  • @KaliMa, for each rational $t$ you find the other intersection point on the circle, call it $(x,y).$ It satisfies $x^2 + y^2 = 5.$ You carefully write $x,y$ with the same denominator, a positive integer, let us call it $q,$ so that $qx$ and $qy$ are integers. Then you have integral tripe $(qx)^2 + (qy)^2 = 5 q^2.$ And those are all. – Will Jagy Dec 03 '12 at 05:01
  • @AdriánBarquero I looked at that already and got as far as (a/c)^2 + (b/c)^2 = 5 but then it immediately seems to skip to the final step without explaining how it got there – KaliMa Dec 03 '12 at 05:04
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    @KaliMa, finish up with this method and you will both have your answer and will have learned something. Gerry is quite able to care for himself...but you don't really get to prescribe the way the (correct) method is described to you. – Will Jagy Dec 03 '12 at 05:09
  • @KaliMa You're right. Look at this derivation instead. It has the necessary details, I think. Also, you can google for "rational parametrization of the unit circle" to find more about this. – Adrián Barquero Dec 03 '12 at 05:10
  • @AdriánBarquero Can I use any one arbitrary point to create the parameterizations? what would you recommend? – KaliMa Dec 03 '12 at 05:14
  • @KaliMa You can use any arbitrary rational point. You will just get different parametrizations if you use different points. You should actually try with a couple of different points to compare the parametrizations that you get. – Adrián Barquero Dec 03 '12 at 05:21
  • @AdriánBarquero I am trying this out now and I am getting some REALLY crazy equations. No idea if I am doing this right. For x I get x = (sqrt(5)sqrt(1-4m^2)-5m^2)/(m^2+1), (-5m^2-sqrt(5)sqrt(1-4m^2))/(m^2+1), and then y is even more crazy – KaliMa Dec 03 '12 at 05:24
  • Square roots? where are they coming from? You have a line through, say, $(1,2)$, with slope $t$. It hits the circle in another point, $(x,y)$. So $(x,y)$ satisfies $x^2+y^2=5$ and $y-2=(x-1)t$. You get a quadratic for, say, $x$, and you know one root of that quadratic is $x=1$, and you know something about the sum and/or product of the two roots of a quadratic, so you get the other one without doing any square roots. Similarly, you get a formula for $y$. So you have $(x(t))^2+(y(t))^2=5$. Multiply everything out, and let $t=m/n$, and clear fractions. – Gerry Myerson Dec 03 '12 at 05:49
  • @GerryMyerson I had solved for x from x^2 + (m(x+5))^2 = 5 – KaliMa Dec 03 '12 at 05:50
  • @GerryMyerson, I wrote it out as a CW answer. Is that alright with you? Easy enough to delete. – Will Jagy Dec 03 '12 at 05:52
  • Kalima, as I wrote, no need for square roots to solve a quadratic when you already know one solution. The two solutions of $ax^2+bx+c=0$ add up to ... ? – Gerry Myerson Dec 03 '12 at 06:31
  • @Will, all is right with me. – Gerry Myerson Dec 03 '12 at 06:32
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Here is a cute way to construct a family of solutions for the given diophantine:

Key Theorem: Let $a, b, c, d \in \mathbb Z$. Then, $$(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$$ Proof: Expand.

Consider a pythagorean triple $(a, b, c)$, which you can easily generate. A simple application of the key theorem yields:

$a^2 + b^2 = c^2$, $\implies$ $5(a^2 + b^2) = 5c^2$, so $(2^2 + 1^2)(a^2 + b^2) = 5c^2$ $\implies$ $(2a + b)^2 + (2b - a)^2 = 5c^2$.

Therefore, if $(a,b,c)$ is a pythagorean triple, then $(2a + b, 2b - a, c)$ is a solution of $x^2 + y^2 = 5z^2$.

$\blacksquare$.

KVS02
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