# Given that $x=\frac{a}{\frac{a}{x+b}+c+b}+c$, prove that $c=-b$

Question: Given that $x=\frac{a}{\frac{a}{x+b}+c+b}+c$ for all real $x\neq -b$, prove that $a\neq 0$ and that $c=-b$.

Proving that $a\neq 0$ probably involves a proof by contradiction, but I am not sure exactly what this involves. I also think this can be shown without proving that $c=-b$, but $a\neq 0$ follows directly if we can prove that first.

To prove that $c=-b$, the first equation can be manipulated to obtain the result $a(b+c)=(x-c)(x+b)(b+c)$. We can safely divide both sides by $b+c$ as if $b+c=0$, $b=-c$ is satisfied anyway. This gets that $a=(x-c)(x+b)$. If we can prove that $a\neq 0$, then it follows that $x-c\neq 0$, and so $x\neq c$. This, because the only original restraint on $x$ was that $x\neq -b$, suggests that $c=-b$, but I am not sure this logic is rigorous or correct.

If someone can prove that $a\neq 0$ and the logic at the end is valid, then I think the problem is solved, unless I made a mistake elsewhere.

Also, this problem is only designed for strong year 10 students, so the solution is unlikely to involve much more than simple proof methods and basic algebraic manipulation.

I shall assume that you want the statement to hold true for all $x$. Suppose on the contrary that $a=0$, then we have $x=c$, but we can let $x$ takes two different values, $x_1 \ne x_2$ and we obtain a contradiction since we will be having $x_1 =c=c_2$. Hence $a \ne 0$.

$$x-c=\frac{a}{\frac{a}{x+b}+b+c}$$

$$x-c=\frac{a(x+b)}{a+(b+c)(x+b)}$$

$$(x-c)[a+(b+c)(x+b)]=a(x+b)$$

Suppose $c \ne -b$, and the only condition on $x$ is $x \ne -b$, then we can let $x=c$.

$$0=a(b+c)$$

Since $a \ne 0$, we have $b+c=0$.

• I can't quite follow your explanation for why $a\neq 0$, but if it can be clarified, then the second part follows well. Thanks. – Jonathan Zhou May 4 '18 at 10:00
• If $a=0$, then you have for all $x$, $x=c$, that is every number that is not equal to $-b$ is equal to $c$, which is a contradiction right. For example if $-b=0$ and $c=1$, you can let $x=2$ and you get $1=2$ which is a contradiction. – Siong Thye Goh May 4 '18 at 16:13
• Thanks, makes sense now. Now it seems we can just assume that $c\neq -b$ so we can let $x=c$, which makes $0=\frac{a}{\frac{a}{x+b}+b+c}$, and so $0=a$, which is a contradiction, therefore our assumption that $c\neq -b$ was false. – Jonathan Zhou May 4 '18 at 23:45

I think you need to take into account the fact that the question asks you to prove that $a \neq 0$ before proving that $c=-b$.

I suppose that $x$ is a variable, which will not take the value $x=-b$...

Let's suppose $a=0$.

Then you have $\forall x, x=c$, which is not possible. Hence $a \neq 0$.

Your initial equation becomes, as you said $a(b+c)=(x-c)(x+b)(b+c)$ by multiplying both sides by $\frac{a}{x+b}+c+b$ and then by $x+b$. Both expressions are not null (remember that $x \neq -b$).

Since $a \neq 0$, I have $(b+c)=\frac 1a(x-c)(x+b)(b+c)$ $\forall x$.

Or $(b+c)(\frac 1a(x-c)(x+b)-1)=0$ $\forall x$

This is only possible if $c=-b$.

• Why is $x=c$ not possible? Also how did you get from the second last line to the final line? – Jonathan Zhou May 4 '18 at 10:24
• @JonathanZhou If $x$ is a variable, it cannot be forced to be $=c$... For the other question, see the edit. – Martigan May 4 '18 at 11:11
• Thanks, this solution works nicely too. – Jonathan Zhou May 4 '18 at 23:47

IMO, the claim is wrong. If $a=0$, the equation reduces to $x=c$, unless $c+b=0$. But we know that $c+b=x+b\ne0$.

For example

$$1=\frac0{\dfrac0{1+1}+1+1}+1.$$

As $x+b\ne0$, we can always rewrite

$$x=\frac{a}{\dfrac{a}{x+b}+b+c}+c=\frac{ax+ab}{(b+c)x+b^2+bc+a}+c\\ =\frac{(a+bc+c^2)x+(b^2c+bc^2+ac+ab)}{(b+c)x+b^2+bc+a}.$$

For this to be an identity, the coefficient of $x$ in the denominator must be zero. When this is the case, it simplifies to

$$x=\frac{ax}{a}.$$

In conclusion, the problem statement is at best ambiguous and at worst wrong. It should say "for all $x\ne-b$".

• Edited, it did mean for all $x\neq -b$. Thanks. – Jonathan Zhou May 4 '18 at 23:36

Note that from

$$x=\frac{a}{\frac{a}{x+b}+c+b}+c$$

we need

• $x+b \neq 0 \implies x\neq -b$

and

• $\frac{a}{x+b}+c+b\neq 0$

then

$$x=\frac{a}{\frac{a}{x+b}+c+b}+c\iff \frac{ax}{x+b}+cx+bx=a+\frac{ac}{x+b}+c^2+bc\\\iff ax+cx^2+bx^2+bcx+b^2x=ax+ab+ac+c^2x+bcx+c^2b+b^2c\\\iff (b+c)x^2+(a+bc+b^2-a-c^2-bc)x-(ab+ac+c^2b+b^2c)=0\\\iff (b+c)x^2+(b^2-c^2)x-(ab+ac+c^2b+b^2c)=0$$

and then for the $x^2$ and $x$ term

• $b=-c$

and for the constant term

• $ab+ac+c^2b+b^2c=ab-ab+b^3-b^3=0$
• Very good your answers. I have upvoted all users. +1 – user401938 May 4 '18 at 9:59
• @Sebastiano Thanks, really appreciate! – user May 4 '18 at 10:01