MathLabs

Problem 5

Let P(x)P(x) be a polynomial of degree n>1n>1 with integer coefficients, and let kk be a positive integer. Define Q(x)=P(P(…P(x)…))Q(x)=P(P(\ldots P(x)\ldots)), where PP occurs kk times. Prove that there are at most nn integers tt such that Q(t)=tQ(t)=t.
Step 6 of 8: Convert mutual divisibility into signs
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α−b=±(β−a),α−a=±(β−b)\alpha-b=\pm(\beta-a),\qquad \alpha-a=\pm(\beta-b)
Detailed analysis

The two mutual divisibilities from Step 5 imply each pair has equal absolute value, hence α−b=±(β−a)\alpha-b=\pm(\beta-a) and α−a=±(β−b)\alpha-a=\pm(\beta-b). If both signs were positive, subtracting the equations would give a−b=b−aa-b=b-a, contradicting a≠ba\ne b. Thus at least one sign is negative, and that equation yields α+β=a+b\alpha+\beta=a+b.