@Pi Han Goh
–
Thank for your help! My friend solves it by \( f(x)=f( \sqrt{x+1}) \)
\( ( x+1)( x^{2}+1)=( \sqrt{x+1}+1)( \sqrt{x+1}^{2}+1) \)
It also is \( f(a)=(a^{2}+1)(a+1) \)
<=> \( x= \sqrt{x+1}
=> x^{2}-x-1=0 \)

@Cong Thanh
–
That's a very nice approach! But the thing is, you didn't prove that they are the only solutions. It make sense that if \(f(x) = f(\sqrt{x+1}) \), then \(x = \sqrt{x+1} \), but does that mean that they are the only solution?

@Pi Han Goh
–
This comes from the one-one nature of \(f(x)\).

\[f(x)=x^3+x^2+x+1\]
Since \(f(x)\) is a odd degree polynomial, for proving \(f(x)\) to be one-one, it suffices to prove \(f(x)\) is monotonically increasing.

\[f'(x)=3x^2+2x+1>0\forall x\in\mathfrak R\]
( This comes from the fact that discriminant of \(f'(x)\) is \(-ve\).)

@Rishabh Cool
–
No no. You're missing the point here. the domain for \(x\) is \(x>-1\) right? Just show that \(3x^2 + 2x + 1\) is strictly positive and you're done.

Of course you can use quad discriminant as well....

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## Comments

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TopNewestWhat have you tried? Have you tried squaring the equation? The only real solution is \( \dfrac{1+\sqrt5}2 \).

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Can you solve it in steps? It also has \( (1- \sqrt{5})/2 \)

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Oh right, I forgot to count the negative sign.

Rearrange the equation: \( \dfrac{(x+1)(x^2+1)}{x+2} - 1 = \sqrt{x+1} \). Now square both sides of the equation and simplify it.

You will get the equation \(x^6+2 x^5+x^4-3 x^3-7 x^2-8 x-3 = 0 \). Factorize it and show that it only has two real roots: \( \dfrac{1\pm\sqrt5}2 \).

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\[f(x)=x^3+x^2+x+1\] Since \(f(x)\) is a odd degree polynomial, for proving \(f(x)\) to be one-one, it suffices to prove \(f(x)\) is monotonically increasing.

\[f'(x)=3x^2+2x+1>0\forall x\in\mathfrak R\] ( This comes from the fact that discriminant of \(f'(x)\) is \(-ve\).)

Hence \(f(x_1)=f(x_2)\iff x_1=x_2\forall x_1,x_2\in\mathfrak R\).

Is it right @Pi Han Goh ?

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Of course you can use quad discriminant as well....

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