[{"_1":2,"_3":-5,"_4":-5},"loaderData",{"_5":6,"_7":8},"actionData","errors","root",{"_169":170,"_171":60,"_172":15,"_173":15,"_174":15,"_175":15,"_176":15,"_177":15,"_178":179},"routes/s.$postId",{"_9":10,"_11":12,"_13":-7,"_14":15,"_16":15,"_17":18},"kind","post_with_profile","postWithProfile",{"_19":20,"_21":22},"referrer","preferIntrinsicAspectRatioForOgImage",false,"enforceRenderedOgImg","defaultOgDescription","ChatGPT helps you get answers, find inspiration, and be more productive.","post",{"_41":42,"_43":44,"_45":46,"_47":48,"_49":50,"_51":52,"_53":54,"_34":55,"_56":57,"_58":15},"profile",{"_23":24,"_25":24,"_26":27,"_28":-5,"_29":-5,"_30":31,"_32":15,"_33":-5,"_34":-5,"_35":-5,"_36":15,"_37":-5,"_38":-5,"_39":-5,"_40":15},"user_id","anonymous","username","profile_picture_url","https://cdn.openai.com/sora/images/profile_placeholder_v4.png","profile_picture_id","is_default_profile_picture","follower_count",0,"is_following","display_name","permalink","can_cameo","verified","character_count","owner_profile","social_context","follows_you","id","t_69f02c082500819188341babb6cb86e8","posted_at",1777347592.144663,"text","126: 1","og_title","See what this chat’s about","og_description","Open this chat to take a look.","attachments",[66],"preview_image_url","https://ogimg.chatgpt.com/?postId=t_69f02c082500819188341babb6cb86e8","https://chatgpt.com/s/t_69f02c082500819188341babb6cb86e8","permissions",{"_59":60,"_61":15,"_62":15,"_63":60,"_64":65},"use_experimental_unfurl","can_read",true,"can_write","can_delete","can_remix","share_setting","public",{"_41":67,"_68":69,"_9":70,"_71":72,"_73":-5},"t_69f02c082500819188341babb6cb86e8-attachment-0","tags",[],"message_slice","messages",[74,75],"group_chat_id",{"_41":125,"_77":136,"_79":137,"_81":138,"_82":139,"_84":85,"_86":-5,"_87":31,"_89":140,"_91":92,"_93":-5},{"_41":76,"_77":78,"_79":80,"_81":-5,"_82":83,"_84":85,"_86":60,"_87":88,"_89":90,"_91":92,"_93":94},"c5fc5560-a844-4138-a8b7-17af21710d60","author",{"_132":133,"_134":-5,"_89":135},"create_time",1777347294.647451,"update_time","content",{"_128":45,"_129":130},"status","finished_successfully","end_turn","weight",1,"metadata",{"_95":96,"_97":98,"_99":100,"_101":102,"_103":104,"_105":106,"_107":108,"_109":110,"_111":112,"_113":114,"_115":100,"_116":100,"_117":118,"_119":60,"_120":121,"_122":123,"_124":125},"recipient","all","channel","final","citations",[],"content_references",[],"resolved_model_slug","gpt-5-5-pro","search_result_groups",[],"request_id","wfr_019dd220102f76c398e28657801035f7","message_type","next","turn_exchange_id","cacadb92-e62f-4d0e-8dc6-954421b5efce","poll_on_websocket_inactivity_ms",30000,"poll_freshness_max_mins",120,"poll_interval_ms",10000,"model_slug","default_model_slug","thinking_effort","extended","can_save","targeted_feedback",{"_126":127},"async_source","saserver-centralus-prod.fck9d:bon-user-VIeSd3gKmElvaLpglzpGXuXd-33e77119-0616-4ff7-863c-67d32a145d80:US","parent_id","4b3de612-a35c-4f00-baa4-20cd74839d4a","category","pro","content_type","parts",[131],"## Definitions\n\nAll logarithms are natural. For a finite nonempty set \\(A\\subset\\mathbb N\\), define\n\n\\[\nP(A):=\\prod_{\\substack{a,b\\in A\\\\a\\ne b}}(a+b),\n\\qquad\nk(A):=\\omega(P(A)),\n\\]\n\nwhere \\(\\omega(m)\\) is the number of distinct prime divisors of \\(m\\). For \\(|A|=1\\), take \\(P(A)=1\\), so \\(k(A)=0\\).\n\nLet\n\n\\[\nH(A):=\\frac{\\max A}{\\gcd(A)}.\n\\]\n\nFor a finite set \\(S\\) of primes, write\n\n\\[\nU_S:=\\{m\\in\\mathbb N:\\text{ every prime divisor of }m\\text{ lies in }S\\}.\n\\]\n\nFor \\(m\\in\\mathbb N\\), write\n\n\\[\nm_S:=\\prod_{p\\in S}p^{\\nu_p(m)}.\n\\]\n\nThus \\(m_S\\) is the \\(S\\)-part of \\(m\\).\n\n---\n\n# Theorem 1: height-dependent determinant bound\n\nThere is an absolute constant \\(C>0\\) such that for every finite nonempty \\(A\\subset\\mathbb N\\),\n\n\\[\n\\boxed{\n|A|\n\\le\nC\\,(k(A)+1)\\,(1+\\log H(A))\\,(1+\\log\\log(3H(A))).\n}\n\\]\n\n---\n\n## Lemma 1: primitive normalization\n\nLet \\(A\\subset\\mathbb N\\) be finite with \\(|A|\\ge2\\), and let \\(S\\) be the set of primes dividing \\(P(A)\\). Put\n\n\\[\ng:=\\gcd(A),\n\\qquad\nA^\\ast:=g^{-1}A.\n\\]\n\nThen \\(A^\\ast\\subset\\mathbb N\\), \\(\\gcd(A^\\ast)=1\\), \\(|A^\\ast|=|A|\\), \\(H(A^\\ast)=H(A)\\), and every off-diagonal sum in \\(A^\\ast\\) lies in \\(U_S\\).\n\n### Proof\n\nFor distinct \\(a,b\\in A\\), the integer \\(a+b\\) lies in \\(U_S\\). Since \\(g\\mid a\\) and \\(g\\mid b\\), we have \\(g\\mid a+b\\). Hence every prime divisor of \\(g\\) belongs to \\(S\\). Therefore \\(g\\in U_S\\).\n\nFor distinct \\(a,b\\in A\\),\n\n\\[\n\\frac{a}{g}+\\frac{b}{g}=\\frac{a+b}{g}.\n\\]\n\nSince \\(a+b\\in U_S\\) and \\(g\\in U_S\\), the quotient \\((a+b)/g\\) also belongs to \\(U_S\\). The remaining claims are immediate from the definition of \\(A^\\ast\\). \\(\\square\\)\n\n---\n\n## Lemma 2: Cauchy determinant deletion lemma\n\nLet \\(S\\) be a finite set of primes with \\(|S|=k\\). Let\n\n\\[\nX=\\{x_1,\\dots,x_r\\},\n\\qquad\nY=\\{y_1,\\dots,y_r\\}\n\\]\n\nbe two sets of distinct positive integers such that\n\n\\[\nx_i+y_j\\in U_S\n\\qquad\n(1\\le i,j\\le r).\n\\]\n\nDefine\n\n\\[\n\\Delta(X):=\\prod_{1\\le i0\\). Therefore\n\n\\[\nQ\\det\\mathcal C\\in\\mathbb Z\\setminus\\{0\\},\n\\]\n\nso\n\n\\[\n|\\det\\mathcal C|\\ge \\frac1Q.\n\\]\n\nThe Cauchy determinant identity gives\n\n\\[\n|\\det\\mathcal C|\n=\n\\frac{\\Delta(X)\\Delta(Y)}\n{\\prod_{i,j}(x_i+y_j)}.\n\\]\n\nTherefore\n\n\\[\n\\frac{\\prod_{i,j}(x_i+y_j)}{Q}\n\\le\n\\Delta(X)\\Delta(Y).\n\\tag{1}\n\\]\n\nNow set\n\n\\[\nU:=\\{(i,\\pi_p(i)):\\ p\\in S,\\ 1\\le i\\le r\\}.\n\\]\n\nThis is a union of at most \\(k\\) matchings.\n\nLet\n\n\\[\nP:=\\prod_{i,j}(x_i+y_j),\n\\qquad\nP_{\\bar U}:=\\prod_{(i,j)\\notin U}(x_i+y_j).\n\\]\n\nFor each \\(p\\in S\\),\n\n\\[\n\\nu_p(P_{\\bar U})\n=\n\\sum_{(i,j)\\notin U}e_{ij}^{(p)}\n\\le\n\\sum_{\\substack{1\\le i,j\\le r\\\\ j\\ne \\pi_p(i)}}e_{ij}^{(p)}\n=\n\\nu_p(P)-M_p\n=\n\\nu_p(P/Q).\n\\]\n\nFor primes \\(q\\notin S\\), both \\(P_{\\bar U}\\) and \\(P/Q\\) have \\(q\\)-adic valuation \\(0\\). Hence\n\n\\[\nP_{\\bar U}\\mid P/Q.\n\\]\n\nCombining this with (1) gives\n\n\\[\nP_{\\bar U}\\le \\Delta(X)\\Delta(Y).\n\\]\n\n\\(\\square\\)\n\n---\n\n## Lemma 3: column-sensitive inequality for separated blocks\n\nLet \\(S\\) be a finite set of primes with \\(|S|=k\\). Let\n\n\\[\nX=\\{x_1<\\cdotsk\\), then\n\n\\[\n\\boxed{\n\\prod_{j=1}^r y_j^{\\,r-k}\n\\le\n\\Delta(X)\\Delta(Y).\n}\n\\tag{2}\n\\]\n\nConsequently, if\n\n\\[\nD(X):=x_r-x_1,\n\\qquad\nD(Y):=y_r-y_1,\n\\]\n\nthen\n\n\\[\n\\boxed{\nD(X)D(Y)\n\\ge\ny_1^{\\,2(r-k)/(r-1)}.\n}\n\\tag{3}\n\\]\n\n### Proof\n\nApply Lemma 2. We obtain a set \\(U\\subseteq [r]\\times[r]\\), a union of at most \\(k\\) matchings, such that\n\n\\[\n\\prod_{(i,j)\\notin U}(x_i+y_j)\n\\le\n\\Delta(X)\\Delta(Y).\n\\tag{4}\n\\]\n\nFix a column \\(j\\). Since \\(U\\) is a union of at most \\(k\\) matchings, at most \\(k\\) pairs in column \\(j\\) belong to \\(U\\). Therefore at least \\(r-k\\) pairs \\((i,j)\\) do not belong to \\(U\\). For each such pair,\n\n\\[\nx_i+y_j\\ge y_j.\n\\]\n\nThus\n\n\\[\n\\prod_{(i,j)\\notin U}(x_i+y_j)\n\\ge\n\\prod_{j=1}^r y_j^{r-k}.\n\\]\n\nTogether with (4), this proves (2).\n\nFor (3), observe that\n\n\\[\n\\prod_{j=1}^r y_j^{r-k}\\ge y_1^{r(r-k)}.\n\\]\n\nAlso,\n\n\\[\n\\Delta(X)\\le D(X)^{r(r-1)/2},\n\\qquad\n\\Delta(Y)\\le D(Y)^{r(r-1)/2}.\n\\]\n\nHence (2) gives\n\n\\[\ny_1^{r(r-k)}\n\\le\n\\bigl(D(X)D(Y)\\bigr)^{r(r-1)/2}.\n\\]\n\nTaking powers \\(2/[r(r-1)]\\) yields\n\n\\[\nD(X)D(Y)\n\\ge\ny_1^{2(r-k)/(r-1)}.\n\\]\n\n\\(\\square\\)\n\n---\n\n## Lemma 4: growth lemma\n\nThere is an absolute constant \\(C_0>0\\) with the following property.\n\nLet \\(H\\ge1\\), and let\n\n\\[\n1\\le y_1k\\). Indeed, if \\(k=1\\), then \\(r=2>1\\). If \\(k\\ge2\\), then\n\n\\[\nr-1\\ge (k-1)L+1>k-1.\n\\]\n\nAlso,\n\n\\[\nr\\le (k-1)L+3\\le (k+1)L\n\\]\n\nbecause \\(L\\ge8\\).\n\nIf \\(n<2r\\), then\n\n\\[\nn<2r\\le 2(k+1)L\n\\le 16(k+1)(1+\\log H),\n\\]\n\nand the desired bound follows because \\(1+\\log\\log(3H)>0\\). Hence assume\n\n\\[\nn\\ge2r.\n\\]\n\nOrder the elements of \\(A\\) as\n\n\\[\nA=\\{a_10\\).\n\nWrite\n\n\\[\nu:=\\nu_p(x),\n\\qquad\nv:=\\nu_p(c).\n\\]\n\nIf \\(u\\ne v\\), then\n\n\\[\n\\nu_p(x+c)=\\min(u,v),\n\\]\n\nso\n\n\\[\n\\lambda_p(c)=0,\n\\]\n\ncontrary to assumption. Therefore\n\n\\[\n\\nu_p(x)=\\nu_p(c)=:t.\n\\]\n\nSince\n\n\\[\n\\lambda_p(c)=\\nu_p(x+c)-t,\n\\]\n\nwe have\n\n\\[\n\\frac{x}{p^t}+\\frac{c}{p^t}\n\\equiv 0\n\\pmod {p^{\\lambda_p(c)}}.\n\\tag{12}\n\\]\n\nBy the choice of \\(c_p\\),\n\n\\[\n\\lambda_p(c_p)\\ge \\lambda_p(c)>0.\n\\]\n\nThe same argument gives\n\n\\[\n\\nu_p(c_p)=\\nu_p(x)=t,\n\\]\n\nand\n\n\\[\n\\frac{x}{p^t}+\\frac{c_p}{p^t}\n\\equiv 0\n\\pmod {p^{\\lambda_p(c)}}.\n\\tag{13}\n\\]\n\nSubtracting (13) from (12),\n\n\\[\n\\frac{c-c_p}{p^t}\n\\equiv 0\n\\pmod {p^{\\lambda_p(c)}}.\n\\]\n\nThus\n\n\\[\n\\nu_p(c-c_p)\\ge t+\\lambda_p(c).\n\\]\n\nSince\n\n\\[\n\\nu_p(c)=\\nu_p(c_p)=t,\n\\]\n\nwe get\n\n\\[\n\\nu_p\\left(\\frac{|c-c_p|}{\\gcd(c,c_p)}\\right)\n=\n\\nu_p(c-c_p)-t\n\\ge\n\\lambda_p(c).\n\\]\n\nThis proves (11).\n\nBecause \\(c\\notin B_x\\), we have \\(c\\ne c_p\\) for every \\(p\\in S\\), so the factor corresponding to \\(d=c_p\\) occurs in the product defining \\(R_C(c)\\). Therefore, for every \\(p\\in S\\),\n\n\\[\n\\nu_p\\left(\\frac{x+c}{\\gcd(x,c)}\\right)\n\\le\n\\nu_p\\left(\n\\prod_{\\substack{d\\in C\\\\d\\ne c}}\n\\left(\\frac{|c-d|}{\\gcd(c,d)}\\right)_S\n\\right).\n\\]\n\nSince \\((x+c)/\\gcd(x,c)\\in U_S\\), this implies\n\n\\[\n\\frac{x+c}{\\gcd(x,c)}\n\\mid\n\\prod_{\\substack{d\\in C\\\\d\\ne c}}\n\\left(\\frac{|c-d|}{\\gcd(c,d)}\\right)_S.\n\\tag{14}\n\\]\n\nMultiplying (14) by (10), we obtain\n\n\\[\nx+c\\mid\nc_S\n\\prod_{\\substack{d\\in C\\\\d\\ne c}}\n\\left(\\frac{|c-d|}{\\gcd(c,d)}\\right)_S\n=\nR_C(c).\n\\]\n\nThus every \\(c\\in C\\setminus B_x\\) is good. Since \\(|B_x|\\le k\\), the set\n\n\\[\nG_x:=C\\setminus B_x\n\\]\n\nhas size at least \\(|C|-k\\). \\(\\square\\)\n\n---\n\n## Proof of Theorem 2\n\nLet \\(C\\subset A\\) have size \\(k+1\\), and set\n\n\\[\nM:=\\max C.\n\\]\n\nIf \\(A=C\\), then\n\n\\[\n\\max A=M\\le M^{k+1},\n\\]\n\nso assume \\(x\\in A\\setminus C\\).\n\nBy Lemma 5, since \\(|C|=k+1\\), there exists \\(c\\in C\\) such that\n\n\\[\nx+c\\mid R_C(c).\n\\]\n\nWe bound \\(R_C(c)\\). First,\n\n\\[\nc_S\\le c\\le M.\n\\]\n\nFor each \\(d\\in C\\), \\(d\\ne c\\),\n\n\\[\n\\left(\\frac{|c-d|}{\\gcd(c,d)}\\right)_S\n\\le\n\\frac{|c-d|}{\\gcd(c,d)}\n\\le\n|c-d|\n\\le\nM.\n\\]\n\nThere are exactly \\(k\\) factors in the product over \\(d\\ne c\\), because \\(|C|=k+1\\). Hence\n\n\\[\nR_C(c)\\le M^{k+1}.\n\\]\n\nSince \\(x+c\\mid R_C(c)\\), we have\n\n\\[\nx+c\\le R_C(c)\\le M^{k+1}.\n\\]\n\nTherefore\n\n\\[\nx

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