Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-76/5/ii/solution

Define the normalized progression average
We first prove the needed four-term progression bound with torsion factors, valid for the arbitrary cyclic modulus in the question:
For the factor at slope zero, write , , and apply Cauchy-Schwarz to discard :
For concise bookkeeping put , the conjugate of the preceding derivative convention, and . Expanding the square with and setting gives . Another Cauchy-Schwarz, now in , discards and yields
Expand with , set , and apply Cauchy-Schwarz a third time to discard . The remaining -average is a full average over , so
Negating does not change the average. The nonnegative integrand depends only on and . Averaging over the subgroups and costs at most their indices and compared with unrestricted increments. The unrestricted expression is exactly , obtained by expanding the last squared average. Thus the bound is .
Reversing the progression handles slope three. To isolate slope one, eliminate slopes three, zero and two successively in the same calculation; the surviving increments have coefficients , so only the index is needed. Reversal handles slope two. This proves the minimum-over-factors bound, without assuming that or is invertible modulo .
If , the requested lower bound is zero and is immediate. Otherwise let . Telescoping the product of four copies of against the constant product gives four error terms, each with one factor and three factors bounded by one. Hence
In particular, choosing gives
These are ordered modular progression parameters , as requested; small-order differences and are included. A claim of distinct nontrivial progressions would need a separate subtraction and size condition.

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