Proposition 35 ([Lcm bound]). Suppose f(v)≤φ(v) , g(w)≤φ(w) , and e(v,w)≤f(v)g(w) h(v/(v,w))h(w/(v,w)). If every positive edge satisfies [v,w]≤Z , then, for each fixed 1/2<u<1 , E Z²(1-u) (FG)^u. More generally, in a cell fixing the …

The cases FG=0 are immediate. Fixed-exponent stripping. Fixing exponents at a prime and stripping its powers divides the lcm cutoff by p^ (i,j) . Divide the two vertex weights by φ(p^i),φ(p^j) and the edge weight by their product and by h_p^ 1_ i≠j . The stripped graph meets the same hypotheses. Rescaling a proposed bound costs at most h_p^ 1_ i≠j p^ -2(1-u)(i,j) (φ(p^i)φ(p^j))^ 1-u ≤h_p^ 1_ i≠j p^ -(1-u)|i-j| . Iteration proves the asserted cell coefficient once the main bound is known. It suffices first to prove the bound for graphs with no prime below a fixed large cutoff p_0 . Indeed, after that proof, strip the finitely many smaller primes and sum their exponent cells. Each corresponding kernel h_p^ 1_ i≠j p^ -(1-u)|i-j| has a finite convolution bound on the u -powers of probability vectors. Their finite product only changes the constant. Large-prime induction and common pivots. For the large-prime graph use induction on the number of occurring primes, with a constant to be fixed below. The zero-prime graph is immediate. If the assertion with this constant fails, normalize the edge weights to a probability law. At each occurring prime the induction hypothesis on a fixed exponent cell gives m_ ij ≤p^ -(1-u)|i-j| h_p^ 1_ i≠j (α_iβ_j)^u, where α_i,β_j are the fractions of the two vertex totals in those exponent classes. Lemma 6 ([One-coordinate concentration]) gives a common exponent k_p with total off-pivot vertex mass O(1/p) . The edge probability of two off-pivot exponents is O(p^ -2u ) , and that of a single deviation of size at least two is O(p^ -2(1-u)-u )=O(p^ -(2-u) ). Both sums converge. Increasing p_0 therefore permits us to discard less than half the total edge weight while excluding both events at every prime. Put N=∏_pp^ k_p . Every remaining edge has the form v=Na/b, w=Nc/d_1, [v,w]=Nac, where b,d_1 N and a,b,c,d_1 are pairwise coprime squarefree integers. These statements are edgewise: at a prime at most one side differs from the pivot, and its difference is exactly one. Since (a,b)=(c,d_1)=1 , each vertex has a unique reduced ratio to N ; this remains true when a or c shares primes with N . Moreover, (v,w)=N/(bd_1), h(v/(v,w))h(w/(v,w))=h(ab)h(cd_1). Dyadic capacities and summation. For a∈[A,2A) , set C= Z NA , D_1=NA^2, D_2=NC^2. There are no such edges unless C≥1 . For each fixed m>0 , the total upper vertex weight on the first side, inflated by h(ab)^m , is O_m(D_1) . To see this, use φ(Na/b)≤aφ(N/b), ∑_ b N φ(N/b)N h(b)^m ≤∏_ p N (1+(h_p^m-1)/(p)), and expand h(a)^m into its nonnegative squarefree divisor sum. Divisibility by j among a≤2A occurs at most 2A/j times. Equation proves the bound. The same calculation on c≤C gives O_m(D_2) . Holder's inequality now bounds the retained edges in this dyad by O_m((D_1D_2)^ 1/m ((F,D_1)(G,D_2))^ 1-1/m ). Indeed apply Holder under the actual product measure f(v)g(w) . Its mass on the dyad is at most (F,O(D_1))(G,O(D_2)) , and the m th inflation moment under that measure is O_m(D_1D_2) by the totient capacities. These constants depend only on the prescribed moment bounds, not on N or its prime support. Choose 1-1/m>u . Since D_1D_2=Z^2 , divide this expression by Z^ 2(1-u) (FG)^u and put f_1=F/D_1 , g_1=G/D_2 . The quotient is at most a constant times (f_1,1)^ 1-1/m f_1^u (g_1,1)^ 1-1/m g_1^u ≤(f_1,f_1^ -1 )^ c(u) for some c(u)>0 . This is summable over dyadic A , since D_1=NA^2 changes geometrically. Thus the retained edges satisfy with a constant independent of the number of primes. Choose the inductive constant larger than twice this constant. Since at least half the original edge weight was retained, this contradicts the supposed failure and closes the induction. Stripping small primes as above proves the unrestricted bound and its cell form.