function composition with quadratics yields no real roots (Auckland MO 2024 P11)

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function composition with quadratics yields no real roots (Auckland MO 2024 P11)

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#1 h Mar 12, 2025, 12:34 AM

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It is known that for quadratic polynomials $P(x)=x^2+ax+b$ and $Q(x)=x^2+cx+d$ the equation $P(Q(x))=Q(P(x))$ does not have real roots. Prove that $b \neq d$ .

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ohiorizzler1434

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#2 h Mar 12, 2025, 1:21 AM

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bruh!! It's sufficient to show that if b=d, then there could be real roots! But if we let P(x)=x^2+2x+c, Q(x)=x^2+x+c, we can do casework on the sign of c to find solutions to P(Q(x))=Q(P(x)) using intermediate value theorem.

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#4 h Mar 12, 2025, 9:57 AM

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I will write $PQ$ for the composition.
Note that $PQ(x)=x^4+2cx^3+{\rm lot}$ and that $QP(x)=x^4+2ax^3+{\rm lot}$ .
Hence if $a\not=c$ then $PQ-QP$ has exact degree 3 and it has a real root, contradiction.
It follows that $a=c$ . Then it is clear that $b\not=d$ , otherwise we have $P=Q$ and $PQ=QP$ .

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