MathLabs

Problem 2

For a polynomial PP and a positive integer nn, define PnP_n as the number of positive integer pairs (a,b)(a,b) such that a<b≤na<b\le n and ∣P(a)∣−∣P(b)∣|P(a)|-|P(b)| is divisible by nn. Determine all polynomials PP with integer coefficients such that Pn≤2021P_n\le2021 for all positive integers nn.
Step 5 of 6: For c=1, only exact equality of |P(a)| and |P(b)| matters
∣P(a)∣−∣P(b)∣<n always, so equality ∣a+d∣=∣b+d∣ is forced|P(a)|-|P(b)|<n\ \text{always, so equality }|a+d|=|b+d|\text{ is forced}
Detailed analysis

Take P(x)=x+dP(x)=x+d. If a+d,b+da+d,b+d have the same sign, ∣P(a)∣−∣P(b)∣=±(a−b)|P(a)|-|P(b)|=\pm(a-b) with 0<∣a−b∣<n0<|a-b|<n (as 1≤a<b≤n1\le a<b\le n), so nn never divides it, giving no pairs. If they have opposite signs, say a<−d≤ba<-d\le b (requiring −d=:m>0-d=:m>0), then ∣P(a)∣−∣P(b)∣=−(a+b+2d)|P(a)|-|P(b)|=-(a+b+2d), and since 1≤a≤m−11\le a\le m-1, m≤b≤nm\le b\le n, one checks ∣a+b−2m∣<m≤n|a+b-2m|<m\le n; so again the difference has absolute value less than nn, forcing a+b=2ma+b=2m exactly for nn to divide it.