Let f(n) denote the number of integer solutions of the equation 3x2+2xy+3y2=n
How can one evaluate the limit limn→∞f(1)+...f(n)n
Thanks
Answer
Let us elaborate the solution by Thomas Andrews. Though the idea is simple and repeatedly used in other answers, here we want to give a meticulously detailed solution.
As we have observed, the number f(1)+⋯+f(n) is equal to the integer solution (x,y), or equivalently the number of integer points (x,y)∈Z2, of the inequality
3x2+2xy+3y2≤n.
Now let
Sn={(x,y)∈Z2:(x,y) satisfies (1)}
be the set of integer solutions (x,y) of (1), and let
En={(x,y)∈R2:(x,y) satisfies (1)}
be the set of points (x,y)∈R2 satisfying (1). Plainly, Sn is the set of integer points contained in En and in this notation we have
#(Sn)=f(1)+⋯+f(n),
where # denotes the cardinality of a set. We also define
An={(x,y)∈Zn:C(x,y)⊂En},
and
Bn={(x,y)∈Zn:C(x,y)∩En≠∅},
Now for each integer point p=(x,y)∈Z2, we associate a unit square
C(p)=C(x,y)=[x−12,x+12]×[y−12,y+12]
centered at p=(x,y). Then it is plain to observe that
An⊂Sn⊂Bn
Also, by definition it is clear that we have
#(An)=Area(⋃p∈AnC(p))≤Area(En)≤Area(⋃p∈BnC(p))=#(Bn),
Now let us digress to a rudimentary real analysis and prove some basic facts. Let (M,d) be a metric space. For p∈M and ∅≠A⊂M, consider the minimum distance between p and A
dist(p,A)=inf{d(p,q):q∈A}
and correspondingly the ϵ-neighborhood Aϵ of K defined by
Nϵ(A)={q∈M:dist(q,A)<ϵ}.
Then we have the following proposition.
Proposition. If F is closed, ⋂ϵ>0Nϵ(F)=F.
Proof. Let F′=⋂ϵ>0Nϵ(F). Since F⊂Nϵ(F) for each ϵ>0, we have F⊂F′. To show the converse, let p∈F′. Then for each n we have p∈N1/n(F) and hence there exists qn∈F such that d(p,qn)<2n. This shows that qn→p. Since F is closed, the limit must lie in F. Therefore p∈F and hence F′=F. ////
Now let us return to the original problem. For each p∈Bn∖An, we have C(p)∩∂En≠∅. This implies that
C(p)⊂N2(∂En).
But by the scaling and the homogeneity of the polynomial 3x2+2xy+3y2, clearly we have
En=√nE1={(√nx,√ny):(x,y)∈E1}
and accordingly
N2(∂En)=√nN2/√n(∂E1)
This shows that
#(Bn∖An)=Area(⋃p∈Bn∖AnC(p))≤Area(N2(∂En))=Area(√nN2/√n(∂E1))=nArea(N2/√n(∂E1))
Now we are ready.
From (3), we have
0≤#(Bn)−#(An)n≤Area(N2/√n(∂E1)),
which goes to zero as n→∞ by the monotonicity of the Lebesgue measure together with the fact that Area(∂E1)=0. Here we used the proposition above.Thus from (2), we have
Area(E1)=Area(En)n≤#(Bn)n≤#(An)n+#(Bn)−#(An)n≤Area(En)n+o(1)=Area(E1)+o(1)
and therefore
Area(E1)=limn→∞#(An)n=limn→∞#(Sn)n=limn→∞#(Bn)n=Area(E1).
Now it remains to evaluate the area of the ellipse E1. Let
A=(3113)
so that ⟨x,Ax⟩=3x2+2xy+3y2 for x=(x,y). Then by spectral theory for symmetric matrices, there exists a rotation matrix R such that ⟨Rx,DRx⟩ for D=diag(λ1,λ2). Then with a new coordinate (X,Y)=Rx, we have λ1X2+λ2Y2≤1. Therefore the area of the ellipse E1 is given by
Area(E1)=π√λ1λ2=π√detA=π2√2.
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