Use this Polynomial Identity (x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2 to help you create a list of 4 Pythagorean Triples. Here is a list of the ones I have already used to make 6 Pythagorean triples: (4,1), (2,1), (4,3), (7,3), (5,2), (3,1). I need 4 more, any help would be great!

Question

paxpolaris · Answer

$$\left( x^2+y^2 \right)^2=\left( x^2-y^2 \right)^2+\left( 2xy \right)^2$$

paxpolaris · Answer

if we pick: $x=2$, $y=1$  .... we get $5^2=3^2+4^2$ .... or the Pythagorean triplet (3,4,5)

paxpolaris · Answer

pick any 2 positive integers with $x>y$ .... you'll keep getting Pythagorean triplets

paxpolaris · Answer

if we pick: $x=3$ and $y=2$, we get $13^2=5^2+12^2 \to \large\left(5,12,13\right)$
if we pick: $x=4$ and $y=2$, we get $20^2=12^2+16^2 \to \large(12,16,20)$
if we pick: $x=5$ and $y=3$, we get $34^2=16^2+30^2 \to \large(16,30,34)$
if we pick: $x=6$ and $y=1$, we get $37^2=35^2+12^2 \to \large(12,35,37)$

anonymous · Answer

omg thank you so much! :)