Waste less time on Facebook — follow Brilliant.
×

What a cliché!

There exists a 3thogoras Theorem where......

\((a^{2}+b^{2}+c^{2}=d^{2}\)).

Now: How many solutions are there for triples (\(a,b,c\)) for which

\((1<a<b<c<10^{6}\))

sastify the 3thogoras Theorem?

E.g. : (2,3,6) is a sastified 3thogoras triple because \(2^{2}+3^{2}+6^{2}=7^{2}\).

Note by Bryan Lee Shi Yang
2 years ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

what is E ?

Isn't it should be 'd' ?

And Can we Use Polar substitution ?

\(\displaystyle{a=x=r\sin { \theta } \cos { \varphi } \\ b=y=r\sin { \theta } \sin { \varphi } \\ c=z=r\cos { \theta } \\ d=r}\) Deepanshu Gupta · 2 years ago

Log in to reply

@Deepanshu Gupta Polar substitution won't give you integral solutions, which I think he's after. Also, \( 10E - 6 = 10* 10^{-6} = 10^{-5} \) Siddhartha Srivastava · 2 years ago

Log in to reply

10E+6=\(10^{6}\) Bryan Lee Shi Yang · 2 years ago

Log in to reply

For a Primitive Pythagorean quadruple the solutions satisfy the following identity: \[(a^2 +b^2 +c^2 +d^2)^2 = (2ad +2bc)^2 +(2bd-2ac)^2 +(a^2 +b^2 - c^2 -d^2)^2 \] Curtis Clement · 2 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...