×

# 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, 6 months ago

Sort by:

what is E ?

Isn't it should be 'd' ?

And Can we Use Polar substitution ?

$$\displaystyle{a=x=r\sin { \theta } \cos { \varphi } \\ b=y=r\sin { \theta } \sin { \varphi } \\ c=z=r\cos { \theta } \\ d=r}$$ · 2 years, 6 months ago

Polar substitution won't give you integral solutions, which I think he's after. Also, $$10E - 6 = 10* 10^{-6} = 10^{-5}$$ · 2 years, 6 months ago

10E+6=$$10^{6}$$ · 2 years, 6 months ago

For a Primitive Pythagorean quadruple the solutions satisfy the following identity: $(a^2 +b^2 +c^2 +d^2)^2 = (2ad +2bc)^2 +(2bd-2ac)^2 +(a^2 +b^2 - c^2 -d^2)^2$ · 2 years, 6 months ago