compositon function exercise asked by Mayssoune Zein EL Dine in the math group

Exercise:

Find $f\left( 0 \right)$ such that $f\left( f\left( x \right) \right)={{x}^{2}}-x+1$ where $f:\mathbb{R}\to \mathbb{R}$

Solution: Let $f\left( x \right)={{x}^{2}}-x+1$ and we have $f\left( f\left( x \right) \right)={{x}^{2}}-x+1$

$\Leftrightarrow f\left( {{x}^{2}}-x+1 \right)={{f}^{2}}\left( x \right)-f\left( x \right)+1$

For $x=0\Leftrightarrow f\left( 1 \right)={{f}^{2}}\left( 0 \right)-f\left( 0 \right)+1$

For $x=1\Leftrightarrow f\left( 1 \right)={{f}^{2}}\left( 1 \right)-f\left( 1 \right)+1$

So ${{f}^{2}}\left( 0 \right)-{{f}^{2}}\left( 1 \right)-f\left( 0 \right)+f\left( 1 \right)=0\Leftrightarrow \left( f\left( 0 \right)-f\left( 1 \right) \right)\left( f\left( 0 \right)+f\left( 1 \right) \right)-\left( f\left( 0 \right)-f\left( 1 \right) \right)=0$

$\Leftrightarrow \left( f\left( 0 \right)-f\left( 1 \right) \right)\left( f\left( 0 \right)+f\left( 1 \right)-1 \right)=0\Leftrightarrow f\left( 0 \right)=f\left( 1 \right)\,\,\,or\,\,\,f\left( 0 \right)+f\left( 1 \right)=1$

So we have two cases to be considered in this discussion

If $f\left( 0 \right)=f\left( 1 \right)\Leftrightarrow {{f}^{2}}\left( 0 \right)-f\left( 0 \right)+1=f\left( 0 \right)\Leftrightarrow f\left( 0 \right)\left[ f\left( 0 \right)-1 \right]=f\left( 0 \right)-1\Leftrightarrow f\left( 0 \right)=1$

If $f\left( 0 \right)+f\left( 1 \right)=1\Leftrightarrow f\left( 1 \right)=1-f\left( 0 \right)\Leftrightarrow {{f}^{2}}\left( 0 \right)-f\left( 0 \right)+1=1-f\left( 0 \right)\Leftrightarrow f\left( 0 \right)=0$

Now if $f\left( 0 \right)=1$ then $f\left( f\left( 0 \right) \right)={{f}^{2}}\left( 0 \right)-f\left( 0 \right)+1=1-1+1=1\Leftrightarrow f\left( 1 \right)=1$ true

If $f\left( 0 \right)=0\Leftrightarrow f\left( f\left( 0 \right) \right)={{f}^{2}}\left( 0 \right)-f\left( 0 \right)+1=0-0+1=1\Leftrightarrow f\left( 0 \right)=1$ false

Therefore $f\left( 0 \right)=1$

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the idea of solution credit to Trần Hữu Phúc