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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

$\blacksquare$.