<s>[INST] let $\forall x \neq 0 \ \ \ f(x+\frac{1}{x})=x^2-\frac{1}{x^2}$ then find the $f(x)$ 
let $\forall x \neq 0  \ \ \ f(x+\frac{1}{x})=x^2-\frac{1}{x^2}$ then find the $f(x)$


My try :
$$x^2-\frac{1}{x^2}=(x+\frac{1}{x})(x-\frac{1}{x})$$
And $$(x-\frac{1}{x})^2 =(x+\frac{1}{x})^2-4$$
so we have :
$$f(t)=\pm t \sqrt{t^2-4}$$
it is right ?
 [/INST] Such function cannot exist. Take $x=2$, then we have $$f(2+1/2)= 4-1/4= 15/4$$ Now take $x=1/2$, then we have $$f(2+ 1/2)=1/4-4=-15/4$$But $-15/4\neq 15/4$. Hence such function cannot exist. 
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