Determine a numerical no constant function $f$ such that for all $x_1$, $x_2$ in its domaine of definition, the equality $$f\left(\sqrt{x_1^2+(f(x_1))^2}\right) = f\left(\sqrt{x_2^2+(f(x_2))^2}\right)$$ is verified. Can you give some special property of your solution $f$?
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Interesting question! Have you made any progress? – Peter Woolfitt Jun 03 '15 at 17:39
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Let $g(x) = \sqrt{x^2 + f^2(x)}$. We need to show that $f(g(x_1)) = f(g(x_2))$, for all valid $x_1,x_2$. One way to approach that is to make $g$ constant, that is:
$$\sqrt{x^2 + f^2(x)} = c$$ $$x^2 + f^2(x) = c^2$$ $$f(x) = \sqrt{c^2 - x^2}$$
For some contant $c$. That will satisfy your original equation for all $x_1,x_2 \in [-c,c]$.
Barry
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Hello Barry; I think (maybe wrong...?) that your "One way to approach" can be deduced (I mean it is necessary). Besides your answer has a quite special property. Try, please, to see by yourself what is it. – Piquito Jun 03 '15 at 18:52
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I ask about a remarkable property of your function. This is perhaps something hidden. I´ll give you it later if you can´t arrive to see it – Piquito Jun 03 '15 at 19:06
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Oh, no! I do not give the property because I hope now someone else will perceive it. Nothing to do with "external" (continuity, etc), it an intrinsic one. (Sorry for bad English) – Piquito Jun 03 '15 at 19:12
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The domain of f is the set of x such that $\sqrt{x^2 + f^2(x)}\in\mathbb{R}$; besides, since the given equality is true for every $x_1, x_2$ in the domain of f, we must have $\sqrt{x^2 + f^2(x)}$= constant in order to f don't be constant. Hence $f(x) = \sqrt{c^2 - x^2}$ for an arbitrary constant c > 0, as Barry has answered above.
The special property of the function f is that it is its own inverse, $f\circ f(x) = x$ (it is an involution as can be easily proved).
Piquito
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