As in the title. Substituting $y=x$ we get: $$ f(x)^2-x^2=2f(x)-1 $$ after rearranging, we get: $$ f(x)(f(x)-2)=(x+1)(x-1) $$ And I rather cannot assume that e.g. $f(x)=x+1$ and $f(x)-2=x-1$, so what should I do now?
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1If you assume that $f(x) = x+1$, then the equality would hold true? – Varun Iyer Dec 23 '15 at 11:39
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1surely that's a quadratic you can solve to get $f(x)$ ? – Asier Calbet Dec 23 '15 at 11:42
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Oh, and I didn't notice that. Great, thanks! – Dec 23 '15 at 11:43
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Note that in your first step you had the following equation:
$$f(x)^2-x^2=2f(x)-1$$
Instead of factoring, if we rearrange all the terms on one side we get:
$$f(x)^2 - 2f(x) - x^2 + 1 = 0$$
Note we have a quadratic, with $f(x)$ acting like our $x$ and the term $1-x^2$ being our constant term.
Solving, we get:
$$f(x) = \frac{2 \pm \sqrt{4 -4(1-x^2)}}{2}$$ $$ = \frac{2 \pm \sqrt{4 - 4 + 4x^2)}}{2}$$ $$ = \frac{2 \pm \sqrt{4x^2}}{2}$$ $$ = \frac{2 \pm 2x}{2}$$
Therefore, we get $2$ solutions for $f(x)$:
$$f(x) = \frac{2 + 2x}{2} = 1+x$$ $$f(x) = \frac{2 - 2x}{2} = 1-x$$
Galc127
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Varun Iyer
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Consider $g(x)=f(x)-1$. Then we can write the identity as $$ (g(x)+1)(g(y)+1)-xy=g(x)+g(y)+1 $$ or $$ g(x)g(y)=xy $$ I think you can go on from here.
egreg
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HINT:
The equality is equivalent to $$(1-f(x))\cdot(1-f(y)) = x y$$ and with $h(x) = 1-f(x)$ we have $$h(x) h(y) = x y$$
with solutions $h(x) \equiv x$ and $h(x) \equiv -x$, so $f(x) \equiv 1-x$ and $f(x) \equiv 1+x$ are the two solutions.
orangeskid
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