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If $$f(x)=\frac{ax+1}{x-b} \forall x \in\mathbb{R}-b,ab\neq1,a\neq1$$ is a self inverse function such that $$\frac{f(4)}{4}=\frac{f(12)}{12}={f\left(\frac{1+b}{1-a}\right)}$$The question is to find out $a$ and $b$

For a self inverse function $f(f(x))=x$.So I tried to put $f(x)$ in place of $x$ and solve the resulting equation but it didnot helped me .Is there a more logical way to solve this problem in limited time?Any ideas?Thanks.

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    Note that this is intimately related to a [Blashke product](https://en.wikipedia.org/wiki/Blaschke_product) in Complex Analysis, although this is likely not the best way for you to approach this problem. – 2017-02-23

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Is there a more logical way to solve this problem in limited time?

Hint: if you take the premise to mean that there do in fact exist $a,b$ such as $f=f^{-1}$ then, for a shortcut, you can simply plug in $x=0$ and determine that $f\big(f(0)\big)=f(\frac{-1}{b})=0 \implies a=b\,$. Then use the second condition to determine $a\,$ or, as is the case, prove that no solutions exist.

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    But $a=b=0$ doesnot satisfy the condition given. – 2017-02-23
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    @navinstudent Why not? $f(x)=\frac{1}{x}$ is its own inverse since $1 / (1/x) = x$. – 2017-02-23
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    @dxiv $a=b=0$ doesn't satisfy $\frac{f(4)}4 = \frac{f(12)}{12}$. In addition, I don't see how $f(f(1))=1$ implies $a=b=0$, in fact $a=b=2$ also satisfies $f(f(1))=1$ and $f(f(0))=0$. – 2017-02-23
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    I meant the conditions $$\frac{f(4)}{4}=\frac{f(12)}{12}={f(\frac{1+b}{1-a})}$$ – 2017-02-23
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    @navinstudent I misread those as being part of your attempt to solve the problem, not part of the problem itself. In the latter case, there are no solutions, since $a=b=0$ do not satisfy those additional constraints. – 2017-02-23
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    dxiv, a Moebius transformation is the identity map when its matrix is any scalar multiple of the identity. The correct condition is $a=b,$ not necessarily zero – 2017-02-23
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    @WillJagy Indeed. Thank you for the correction. – 2017-02-23
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Write this as

$$\begin{pmatrix} a && 1 \\ 1 && -b \end{pmatrix}\begin{pmatrix}x \\ 1\end{pmatrix}$$

Then this is self inverse iff

$$\begin{pmatrix} a && 1 \\ 1 && -b \end{pmatrix}^2= \lambda I_2$$

i.e.

$$\begin{pmatrix} a^2+1 && a-b \\ a-b && 1+b^2 \end{pmatrix}= \lambda I_2$$

So $a=b$.

The transformation is projective so all we need now is to check $a,b$ so that your conditions follow.

$$12f(4)=4f(12)\iff 3(4a+1)/(4-a)=(12a+1)/(12-a)$$ $$\iff 3(4a+1)(12-a)=(12a+1)(4-a)$$ $$\iff 143a+36=47a+4$$ $$\iff 96a=-32\iff a=-1/3.$$

Finally

$$\displaystyle f\left({1+b\over1-a}\right)=f(1/2)=1$$ $$\ne -1/52=f(4)/4.$$

So no such transformation exists.

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    $a=b=0$ doesn't satisfy the condition $\frac{f(4)}4 = \frac{f(12)}{12}$. – 2017-02-23
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    Adam, a Moebius transformation is the identity when the matrix is scalar, that is, any scalar multiple of the identity. – 2017-02-23
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    @willjagy yeah projective transform. On mobile so cannot effectively edit atm – 2017-02-23
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    Adam, in general, all varieties of cell phone are evil – 2017-02-23
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    @willjagy there we go. Much better. – 2017-02-23
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    if $96 a = -32$ then $a = -1/3$ Good effort, though – 2017-02-23
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    @willjagy I should have taken my own advice and not tried to do everything in my head while typing on mobile and playing Rocket League. Thanks again for checking my massive fails. :-) – 2017-02-23
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    Notice that the other requirement, the one on $f((1+b)/(1-a))$, is not satisfied. – 2017-02-23
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    @fimpellizieri yeah, I figured, but I'm not in an environment where I can do the other computation in my head. I'll edit the ending when I have some quiet time and a computer. You're also welcome to edit in the punchline if you like and don't want to wait – 2017-02-23
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    Your approach via Möbius machinery is much more elegant than mine. I posted an answer below that's more hands on, but if I got it right this time, then there is no solution. – 2017-02-23
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    @fimpellizieri there we go. – 2017-02-23
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    @AdamHughes The other curious thing is that the conditions $ab \ne 1, a \ne 1$ don't make much sense in the given context. I would suspect a transcript error somewhere. – 2017-02-23
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    @dxiv oh absolutely. – 2017-02-23
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Suppose $b\neq 0$. Then $f(0)=-1/b$ and if $f$ is self inverse

$$0=f(-1/b)=\frac{a/b-1}{b+1/b}=\frac{a-b}{b^2+1}.$$

It follows that $a=b$. Then

$$f(f(x))=\frac{a\frac{ax+1}{x-a}+1}{\frac{ax+1}{x-a}-a}=\frac{a^2x+a+(x-a)}{ax+1-a(x-a)}=\frac{a^2x+x}{1+a^2}=x$$

and $f$ is self-inverse. It suffices now to check the remaining conditions. We have that

$$f\left(\frac{1+a}{1-a}\right)=\frac{a\frac{1+a}{1-a}+1}{\frac{1+a}{1-a}-a}=1$$

so it must be that

$$\frac{\frac{4a+1}{4-a}}{4}=\frac{\frac{12a+1}{12-a}}{12}=1$$

But both equations cannot be satisfied simultaneously.

Now, if $b=0$, then $f(x)=a+\frac{1}{x}$ and hence

$$f(f(x))=a+\frac{1}{a+\frac{1}{x}}=a+\frac{x}{ax+1}=\frac{a^2x+x+a}{ax+1}$$

And if $f$ is self inverse

$$a^2x+x+a=ax^2+x\implies ax^2-a^2x-a=0$$

And unless $a=0$, this can be true for at most two distinct values of $x$, so the only option for $f$ to be self inverse is $f(x)=1/x$. Howevr, the condition on $f(4)$ and $f(12)$ is once again not satisfied.

Perhaps it was wrongly transcribed? The following relations do hold in this case:

$$4\cdot f(4)=12\cdot f(12)=f\left(\frac{1+b}{1-a}\right)$$

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    Recheck your calculations on the $x=f(f(x))=\cdots\;$ line. The fraction actually simplifies to $x$, so $f$ *is* in fact an involution for all $a$. – 2017-02-23
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    @dxiv Wow, atrocious miscalculations. Thank you. I'm not sure if I've miscalculated again, but even showing it is an involution I cannot get the other requirements. – 2017-02-23
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    Looks good to me this time around. – 2017-02-23