# Generating Pythagorean triples for $a^2+b^2=5c^2$?

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.

• Dec 5, 2012 at 20:57

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 $$a^2 + b^2 = 5 c^2$$

• $1445 = 5 \cdot 17^2.$ Dec 3, 2012 at 5:58
• My own miscalculation, apologies Dec 3, 2012 at 5:59
• What is a CW answer? Dec 3, 2012 at 6:01
• Country and Western, although in this case, Community Wiki, which means Will doesn't get any points when people vote for his answer. Dec 3, 2012 at 6:33
• @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? Dec 3, 2012 at 20:15

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 $a^2+b^2=5c^2$.

• 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 Dec 3, 2012 at 4:33
• For each $t$, you get a point. For each rational $t$, you get a rational point. Then you clear denominators to get a triple. Dec 3, 2012 at 4:35
• 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 en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple Dec 3, 2012 at 4:47
• @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. Dec 3, 2012 at 5:00
• @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. Dec 3, 2012 at 5:09

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 $$x^2 + y^2 = 5z^2$$.

$$\blacksquare$$.