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.
Generating Pythagorean triples for $a^2+b^2=5c^2$?
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0See http://math.stackexchange.com/questions/225764/integer-solutions-to-x2y2-5z2/225781 – 2012-12-05
2 Answers
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 $$ a^2 + b^2 = 5 c^2 $$
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0$1445 = 5 \cdot 17^2.$ – 2012-12-03
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0My own miscalculation, apologies – 2012-12-03
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0What is a CW answer? – 2012-12-03
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0Country and Western, although in this case, Community Wiki, which means Will doesn't get any points when people vote for his answer. – 2012-12-03
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0@WillJagy Can't something similar be done to the original Pythagorean triple generators, which checks for gcd(m,n)=1 and opposite parity? Does that apply here to for generating primitives? There's no way to generate *only* primitives? – 2012-12-03
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0@KaliMa, it suffices to require $\gcd(q,r) = 1,$ then $q+r$ odd, finally $q + 2 r \neq 0 \pmod 5.$ – 2012-12-03
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 $a^2+b^2=5c^2$.
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0I 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 – 2012-12-03
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0For each $t$, you get a point. For each *rational* $t$, you get a *rational* point. Then you clear denominators to get a triple. – 2012-12-03
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0I 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 – 2012-12-03
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1@KaliMa In the same link that you give, if you scroll down a little to the section [Geometry of Euclid's Formula](http://en.wikipedia.org/wiki/Pythagorean_triple#Geometry_of_Euclid.27s_formula) you'll find precisely what Gerry is trying to explain to you in the case of the usual Pythagorean Triples. – 2012-12-03
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0@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. – 2012-12-03
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0@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 – 2012-12-03
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2@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. – 2012-12-03
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0@KaliMa You're right. Look at [this derivation](http://mathnow.wordpress.com/2009/11/06/a-rational-parameterization-of-the-unit-circle/) instead. It has the necessary details, I think. Also, you can google for "rational parametrization of the unit circle" to find more about this. – 2012-12-03
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0@AdriánBarquero Can I use any one arbitrary point to create the parameterizations? what would you recommend? – 2012-12-03
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0@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. – 2012-12-03
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0@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 – 2012-12-03
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0Square 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. – 2012-12-03
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0@GerryMyerson I had solved for x from x^2 + (m(x+5))^2 = 5 – 2012-12-03
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0@GerryMyerson, I wrote it out as a CW answer. Is that alright with you? Easy enough to delete. – 2012-12-03
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0Kalima, 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 ... ? – 2012-12-03
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0@Will, all is right with me. – 2012-12-03