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 $2^{2}+3^{2}+6^{2}=7^{2}$ .

#NumberTheory

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Comments

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 - 6 years, 2 months ago

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

Siddhartha Srivastava - 6 years, 2 months ago

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

Curtis Clement - 6 years, 2 months ago

10E+6= $10^{6}$

Bryan Lee Shi Yang - 6 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$