What is the answer of $x^2+y^2+z^2=w^2$ when $x^2/y^2=y^2/z^2=z^2/x^2$ and $x, y , z, w\in\mathbb{N}$.
what is $x^2+y^2+z^2=w^2$ when $x^2/y^2=y^2/z^2=z^2/x^2$?
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algebra-precalculus
diophantine-equations
1 Answers
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I guess you want to solve $x^2+y^2+z^2 = a^2$ in integers for $a\in\mathbb N$. Your condition tells us that $x=y=z$. So you have $3x^2 = a^2$ and hence there is no solution unless $a=0$ (I am not sure if you allow $\mathbb N$ to contain $0$). On the other hand, if $a=0$ then $x=y=z=0$ and the condition you raised does not make sense.
About the non-zero solution: suppose that $x,y,z\neq0$ then $y=mx$ and $z = nx$. So we have $ \frac1{m^2} =\frac{m^2}{n^2} = n^2 $ from which you obtain $m=n=1$.
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0Glad to help you. – 2011-09-23