Editing Openai/68ec50da-cf00-8005-b5f6-b683506e5853 (section)

=== User: Please don't give an abstract/title. ===
Please don't give an abstract/title. Please organize the proof closer to what I wrote.

Keep various numerical constants in the appendix. Please structure them nicely and only keep the really important ones. Report to 7 digits and format it nicely. 

(1) The proof of the square free lemma as written is incorrect. If we have p^2|(ab+1) where we fix b\in \{1,\ldots,n\} and have a range in {1,\ldots, n\} then note then p<=2n. For p between \sqrt{2n} and 2n we have at most 2 possibilities for each prime so this is negligible. For p between \log\log n and \sqrt{2n} we get 2n/p^2 which summed is also negligible. For p<=\log\log n we can obtain the bound you showed. (The proof is quite analogous to square free numbers for 
(2) The case structure is as follows. Follow it precisely and check the numerics very carefully.
Case 1 - A'' is nonempty and there is 3|a for an element a in A''. 
When now applying the square free result to bound A_7/A_18/... note that we may not use the prime 3. This gives an improved bound of the form (1-\prod_{p\ge 5}(1-1/p^2)) and for A_7/A_18 we may also exclude p = 5. This covers the first half of the first page.
Case 2- A'' is nonempty but there is no 3|a for an element a in A''.
Case 2i (Currently labeled Case 2b)- There is an element in A_70/A_99/A_38/A_251 which is divisible by 3. Then we still get to use that the primes 3/5 do not work for A_7 and A_18. We also win a 2/3 on A* as no element is divisible by 3.
Case 2ii (Currently labeled Case 2a) - Here we have that we get a 2/3 on all the elements in A*/A_70/A_99/A_38/A_251 which gives the bound shown.

Case 3- A* is empty but not A_38 \cup A_251.
Case 3i - If both A_38 and A_251 are not empty we get to use a (1-6/pi^2) win on A_70/A_99/A_38/A_251 and a win of (1-6/pi^2 * 25/24) on A_7/A_18 (here we are using that p=5 cannot occur). This wins. 
Case 3ii - If only 1 occurs. Now we only get to win a (1-6/pi^2) on A_70/A_99 but not on A_38 \cup A_251. We still also win a (1-6/pi^2 * 25/24) on A_7/A_18 . This wins.

Case 4 - This is now with A*/A_38/A_{251} all empty. The analysis is based on A_70 and A_99 either both being occupied or only 1 being occupied.
Case 5 - Now all of A''/A_38/A_{251}/A_70/A_99 are empty. In this case either both A_7/A_18 are occupied. This wins because of a (1-25/24 '' 6/pi^2) on each class. Else only 1 is occupied and then we are trivially done.

This also proves that for N sufficiently large the maximum is achieved by exactly 7 mod 25 or 18 mod 25.

Please turn this into a rigorous manuscript. Please be very careful. Please give nice formatting.