Access to social media where the official platforms are blocked.
I co-founded and developed 0media, beginning with a mobile-first Instagram client. The browser keeps control of the encrypted session while a bounded relay moves opaque traffic only to approved hosts.
I now manage outreach, finances, and the expansion of 0media to more platforms and clients.
Post-training a model for the editorial judgment behind a strong index.
For IndexerLabs, I post-trained a Mistral 3 Large variant on more than 1,000 real-world indexes, using roughly 250 million input tokens.
250Minput tokens
1,000+indexes
A low-latency Polymarket research pipeline built around a retrained Moonshine speech model.
It detects arbitrary spoken phrases, emits immutable trigger events, replays order-book state, and tests whether a signal could have produced a defensible fill.
At 13, I independently designed a lunar-orbit colony for 10,000 residents and 5,000 visitors.
I designed the entire project without a mentor, team, or institutional support. The work covers orbital mechanics, station structure, rotation, an artificial sky, architecture, education, microgravity and zero-gravity spaces, agriculture, and supporting experiments.
Read the Luna-CT research projectTesting with five journalists
An agent for the full workflow of individual journalists and publishers, from research and evidence management through drafting, editing, and publication.
Captain learns from a journalist's edits while keeping sources attached to claims and final judgment with the writer. The goal is to give independent journalists the leverage that large publishers are building internally, without taking away their control or agency.
Expanding access
Extending the first client into a broader access layer for blocked social platforms, while preserving the bounded relay model and keeping readable account traffic out of the proxy.
Open source
A network that automatically traces and validates the ownership of independent media outlets.
Proposed architecture: adapters ingest company registries, filings, mastheads, and disclosures as signed claims. Evidence is stored by content hash. Independent validators attest to or challenge relationships, and a temporal graph resolves ownership paths while retaining provenance, confidence, and disputes.
Graph theory
Published in the September issue of Math Horizons, Mathematical Association of America.
This paper treats the four-digit Kaprekar routine as a functional graph. It replaces 9,900 admissible inputs with 54 digit-difference states, then follows the symmetries and directed paths that funnel every state toward the fixed point 6174.
I conceived, researched, and wrote the paper independently, then took it through publication on my own without external help.
Methods: state reduction, functional graphs, symmetry, finite-case convergence.
Abolitionist history
Forthcoming in the Journal of Juvenilia Studies.
This paper asks how William Roscoe developed an abolitionist conscience within eighteenth-century Liverpool. It reconstructs the role of nature, poetry, family, and universal sympathy through early writings and manuscript evidence.
I wrote it in collaboration with Dr. Wiebracht of Stanford University. I worked at Liverpool Central Library to scan and archive thousands of documents, then used Node.js to graph relationships among texts and poems mathematically.
Methods: archival research, close reading, biographical reconstruction, historical context, network analysis.
Arguments from my discrete mathematics portfolio.
n = a² + b²
A lattice square has area n exactly when n is a sum of two integer squares. One side is the vector (a, b); rotating it to (-b, a) gives an equal, perpendicular lattice side and area a² + b².
Original problem 11
Show that there exists a lattice square with area n, where n is a positive integer, if and only if there exist non-negative integers a and b such that n = a² + b².
Proof
(⇒) Assume there exists a lattice square with area n.
Place one vertex of the square at the origin and let one side of the square be the vector v = (a, b), where a, b ∈ Z.
A side perpendicular to v with the same length is obtained by a 90° rotation: w = (-b, a).
The area of the square equals the area of the parallelogram formed by v and w:
Area = |v × w| = a² + b².
Since the square has area n, we get n = a² + b².
(⇐) Now assume n = a² + b² for some non-negative integers a, b.
Consider the vectors v = (a, b) and w = (-b, a).
By the Pythagorean Theorem, |v|² = a² + b² = |w|², so the two sides have equal length.
Their dot product is v · w = a(-b) + b(a) = 0, so the sides are perpendicular.
Thus these vectors form a lattice square whose area is a² + b² = n.
Therefore, a lattice square of area n exists if and only if n = a² + b².
A = (√3 / 4)s²
A 60° rotation of an integer edge vector introduces √3 into the coordinates of the third vertex. Both coordinates cannot remain integers unless the triangle degenerates.
Original problem 2
Is it possible to construct an equilateral triangle whose vertices are all lattice points? If so, provide an example. If not, give a proof that demonstrates why it is not possible.
Proof
Assume one vertex of the triangle, A, is at the origin (0, 0).
Let a second vertex be B = (a, b), where a and b are integers and not both zero.
To form an equilateral triangle, the third vertex C must be obtained by rotating AB = (a, b) by 60°.
A 60° rotation matrix is:
R₆₀ = [[cos 60°, -sin 60°], [sin 60°, cos 60°]] = [[1/2, -√3/2], [√3/2, 1/2]].
Applying this rotation to (a, b) gives:
C = ((a - b√3) / 2, (a√3 + b) / 2).
For C to be a lattice point, both coordinates must be integers.
However, since √3 is irrational, the expressions (a - b√3) / 2 and (a√3 + b) / 2 cannot both be integers unless a = b = 0, which gives a degenerate triangle.
Therefore, it is impossible to construct a non-degenerate equilateral triangle whose vertices are all lattice points.
A = 2s²(1 + √2)
Pick's theorem forces every lattice polygon to have rational area. A regular octagon with lattice side length would instead have an irrational area containing √2, so such an octagon cannot exist.
Original problem 5
Use Exercise 4 to show that it is not possible to construct a regular lattice octagon (8-gon).
Proof
Let the side length of the regular octagon be s. By Pick's Theorem, the area of any polygon with integer coordinates is either an integer or a half-integer, which means the area must be a rational number.
The area of a regular n-gon is A = (n / 4)s² cot(π / n). For an octagon, n = 8, so A = 2s² cot(π / 8).
From Exercise 4(b), we know tan(π / 4) = 1. Using the double-angle formula for tangent:
tan(π / 4) = 2 tan(π / 8) / (1 - tan²(π / 8)).
Therefore, 1 = 2 tan(π / 8) / (1 - tan²(π / 8)).
Solving gives tan²(π / 8) + 2 tan(π / 8) - 1 = 0, so tan(π / 8) = √2 - 1, taking the positive root.
Thus cot(π / 8) = 1 / (√2 - 1) = √2 + 1.
The area of the octagon is therefore A = 2s²(√2 + 1) = 2s² + 2s²√2.
Since the vertices are lattice points, the distance squared between any two adjacent points, s² = (Δx)² + (Δy)², must be an integer.
This makes 2s²√2 irrational, so the total area A is irrational.
This contradicts Pick's Theorem, meaning a regular lattice octagon cannot exist.
B(P) = Σ gcd(|Δx|, |Δy|)
An edge with displacement (m, n) contains gcd(m, n) - 1 interior lattice points. Applying that primitive-step argument to every edge, then correcting for shared vertices, gives a boundary count for any lattice polygon.
Original problem 31
Let m and n be nonnegative integers. Show that there are exactly gcd(m, n) - 1 lattice points on the line segment between the origin and the point (m, n), not including the endpoints.
Proof of problem 31
Let d = gcd(m, n) and write m = dm′ and n = dn′, where gcd(m′, n′) = 1.
The line segment connecting (0, 0) to (m, n) can be parameterized as (x, y) = (sm, sn) for s ∈ [0, 1].
Substituting our expressions for m and n, we get (x, y) = (sdm′, sdn′).
For (x, y) to be a lattice point, sd must be an integer because gcd(m′, n′) = 1.
Let k = sd. Since s ∈ [0, 1], the possible values for k are integers from 0 to d.
This results in the set of lattice points {(km′, kn′) : k = 0, 1, ..., d}.
To exclude the endpoints (0, 0) and (m, n), we exclude k = 0 and k = d.
The remaining values for k are {1, 2, ..., d - 1}.
The number of such points is d - 1, which is exactly gcd(m, n) - 1.
Original problem 32
Let P be a lattice n-gon with vertices p₁ = (a₁, b₁), ..., pₙ = (aₙ, bₙ). Let dᵢ = gcd(aᵢ₊₁ - aᵢ, bᵢ₊₁ - bᵢ) for i = 1, ..., n - 1, and dₙ = gcd(a₁ - aₙ, b₁ - bₙ). Show that the number of lattice points on the boundary of P is B(P) = Σᵢ₌₁ⁿ dᵢ.
Proof of problem 32
Consider one edge of the polygon pᵢ = (aᵢ, bᵢ) to pᵢ₊₁ = (aᵢ₊₁, bᵢ₊₁).
Set Δx = aᵢ₊₁ - aᵢ and Δy = bᵢ₊₁ - bᵢ.
Let dᵢ = gcd(Δx, Δy). Then Δx = dᵢx′ and Δy = dᵢy′, where gcd(x′, y′) = 1.
Every point on this segment has the form (aᵢ, bᵢ) + t(Δx, Δy), where 0 ≤ t ≤ 1.
A lattice point occurs when both coordinates are integers. Since aᵢ and bᵢ are integers, this means tdᵢx′ ∈ Z and tdᵢy′ ∈ Z.
Because gcd(x′, y′) = 1, this happens if and only if tdᵢ ∈ Z.
Thus t = k / dᵢ for some integer k. The condition 0 ≤ t ≤ 1 gives 0 ≤ k ≤ dᵢ.
Hence the lattice points on this edge are (aᵢ, bᵢ) + (k / dᵢ)(Δx, Δy), for k = 0, 1, ..., dᵢ.
There are exactly dᵢ + 1 lattice points on this edge, including both endpoints.
Now sum over all n edges. The total number counted this way is Σᵢ₌₁ⁿ(dᵢ + 1).
However, each vertex is counted twice because it belongs to two adjacent edges. Since there are n vertices, subtract n:
B(P) = Σᵢ₌₁ⁿ(dᵢ + 1) - n = Σᵢ₌₁ⁿdᵢ + n - n = Σᵢ₌₁ⁿdᵢ.
det(A) = ±1
A 2 by 2 integer matrix has an integer inverse exactly when its determinant is 1 or -1. One direction uses multiplicativity of determinants; the other uses the adjugate formula without introducing denominators.
Original problem 49
Let A be a 2 × 2 matrix with entries in Z. Show that A is invertible over Z if and only if det(A) = ±1. Note that you only need to prove this result here for 2 × 2 matrices, but the same result holds for more general n × n matrices with entries in Z. You are of course welcome to prove the more general result here if you wish.
Proof
Let A = [[a, b], [c, d]], where a, b, c, d ∈ Z.
(⇒) Assume A is invertible over Z.
Taking the determinant of both sides, we get det(AA⁻¹) = det(I).
This gives det(A) det(A⁻¹) = 1.
Since both A and A⁻¹ have integer entries, their determinants must be integers.
Thus det(A) and det(A⁻¹) are integers whose product is 1. The only integers that satisfy this are 1 × 1 and (-1) × (-1).
Therefore, det(A) = ±1.
(⇐) Assume det(A) = ±1.
The standard formula for the inverse of a 2 × 2 matrix is:
A⁻¹ = (1 / det(A)) [[d, -b], [-c, a]].
Since det(A) = ±1, the scalar factor 1 / det(A) is also ±1.
Because a, b, c, d ∈ Z, the matrix [[d, -b], [-c, a]] consists entirely of integers.
Multiplying this matrix by the integer ±1 yields a matrix that still has integer entries.
Therefore, A⁻¹ has entries in Z, proving that A is invertible over Z.
A = I + B / 2 - 1
Dissect a lattice polygon into primitive triangles of area 1 / 2. Counting triangles through vertices and boundary edges converts the triangulation into the exact relationship between area, interior points, and boundary points.
Original problem 58
Prove that the area of a primitive lattice triangle is equal to 1 / 2.
Proof of problem 58
Let T be a primitive lattice triangle with sides v and w.
As shown in Exercise 57, the vectors v and w form a Z-basis for Z².
The area of the parallelogram P spanned by v and w is |det(v, w)|, which is 1 for any Z-basis.
The area of triangle T is exactly half the area of the parallelogram P.
Area(T) = (1 / 2) Area(P) = (1 / 2)(1) = 1 / 2.
Original problem 65
Use Exercise 58 to state and prove an equation that relates the area of P to f.
Proof of problem 65
Theorem 5 states that every lattice polygon P can be dissected into f primitive lattice triangles.
From Exercise 58, we know that the area of every primitive lattice triangle is exactly 1 / 2.
The total area A(P) is the sum of the areas of the f triangles in the dissection.
A(P) = f · (1 / 2).
Thus, A(P) = f / 2.
Original problem 66
Show that f = 2v - eᵦ - 2, where f is the number of primitive triangles, eᵦ is the number of boundary edges, and v is the total number of vertices.
Proof of problem 66
Base case: For a single primitive triangle, f = 1, v = 3, and eᵦ = 3. Plugging these into the right side gives 2(3) - 3 - 2 = 1. Since 1 = 1, the formula holds for the simplest case.
Every time we add a new primitive triangle to the existing shape, we must attach it to at least one existing boundary edge. There are two common ways this happens.
Case 1: Attaching to one existing boundary edge.
We add one new vertex and two new boundary edges. The edge we attached to is no longer a boundary edge, so the change in f is +1, the change in v is +1, and the change in eᵦ is +2 - 1 = +1.
2(v + 1) - (eᵦ + 1) - 2 = (2v - eᵦ - 2) + 2 - 1 = f + 1.
The equality is preserved.
Case 2: Attaching to two existing boundary edges, filling a gap.
We add no new vertices. We add one new boundary edge, but the two edges we covered are no longer boundary edges, so the change in f is +1, the change in v is 0, and the change in eᵦ is +1 - 2 = -1.
2v - (eᵦ - 1) - 2 = (2v - eᵦ - 2) + 1 = f + 1.
The equality is preserved.
Since the formula f = 2v - eᵦ - 2 works for the first triangle and stays true every time we add a new triangle, it must be true for any polygon composed of f primitive triangles.
Original problem 67
Finally, show that A(P) = (1 / 2)B(P) + I(P) - 1.
Proof of problem 67
Let B(P) be the number of lattice points on the boundary and I(P) be the number in the interior.
The total number of vertices v in the triangulation is I(P) + B(P).
The number of boundary edges eᵦ in a primitive triangulation is equal to the number of boundary lattice points B(P).
Substitute these into f = 2(I + B) - B - 2, which simplifies to f = 2I + B - 2.
From Exercise 65, A(P) = f / 2.
A(P) = (2I + B - 2) / 2 = I(P) + (1 / 2)B(P) - 1.
This concludes the proof of Pick's Theorem.
p₂q₁ - p₁q₂ = 1
Successive Farey fractions form a primitive lattice triangle with no extra interior or boundary points. Pick's theorem makes its area 1 / 2, while the determinant formula makes the same area half of p₂q₁ - p₁q₂.
Original problem 91
Suppose that p₁ / q₁ and p₂ / q₂ are two successive terms of Fₙ. In this problem, we will use Pick's Theorem to prove that p₂q₁ - p₁q₂ = 1. Let T be the triangle with vertices (0, 0), (p₁, q₁), and (p₂, q₂).
Proof
Part (a): Any point (x, y) strictly inside T can be expressed as t₁(p₁, q₁) + t₂(p₂, q₂), where t₁, t₂ > 0 and t₁ + t₂ < 1.
The y-coordinate is y = t₁q₁ + t₂q₂ < max(q₁, q₂). Since p₁ / q₁ and p₂ / q₂ are in Fₙ, q₁, q₂ ≤ n, which implies y < n.
If (x, y) were an integer lattice point, x / y would represent a slope strictly between the slopes of (p₁, q₁) and (p₂, q₂): p₁ / q₁ < x / y < p₂ / q₂.
But y < n, so x / y would be a valid Farey fraction in Fₙ strictly between p₁ / q₁ and p₂ / q₂, contradicting the fact that they are successive. Thus I(T) = 0.
Part (b): The boundary of T consists of three segments. The segments from (0, 0) to (p₁, q₁) and (p₂, q₂) contain no internal lattice points because the fractions are fully reduced, meaning gcd(pᵢ, qᵢ) = 1.
Any internal lattice point (x, y) on the segment connecting (p₁, q₁) and (p₂, q₂) would satisfy y ≤ max(q₁, q₂) ≤ n. As in part (a), this would create a fraction x / y between p₁ / q₁ and p₂ / q₂ with denominator y ≤ n, contradicting that the two fractions are successive.
Therefore, the only boundary points are the three vertices, so B(T) = 3.
Part (c): By Pick's Theorem, A = I + B / 2 - 1. Substituting our values gives A(T) = 0 + 3 / 2 - 1 = 1 / 2.
Part (d): Using the determinant formula for a triangle with one vertex at the origin, A = (1 / 2)|x₁y₂ - x₂y₁|.
Substituting our vertices gives A(T) = (1 / 2)|p₁q₂ - p₂q₁|. Since p₂ / q₂ > p₁ / q₁, we know p₂q₁ > p₁q₂, so:
A(T) = (1 / 2)(p₂q₁ - p₁q₂).
Part (e): Equating the two area formulas gives 1 / 2 = (1 / 2)(p₂q₁ - p₁q₂). Multiplying both sides by 2 yields p₂q₁ - p₁q₂ = 1.
|ps - qr| = 1
Comparing the squared distance between two Ford-circle centers with the squared sum of their radii shows that the circles are tangent exactly under the same unimodular condition that characterizes adjacent Farey fractions.
Original problem 104
Prove that the representative Ford circles of two distinct fractions are either tangent at one point or wholly external.
Proof of problem 104
Let the two distinct fractions in lowest terms be p / q and r / s. The centers of their Ford circles are (p / q, 1 / 2q²) and (r / s, 1 / 2s²), with radii R₁ = 1 / 2q² and R₂ = 1 / 2s².
The square of the distance d between their centers is:
d² = (p / q - r / s)² + (1 / 2q² - 1 / 2s²)².
The square of the sum of their radii is:
(R₁ + R₂)² = (1 / 2q² + 1 / 2s²)².
We want to find the difference between the squared distance and the squared sum of the radii:
d² - (R₁ + R₂)² = (p / q - r / s)² + (1 / 2q² - 1 / 2s²)² - (1 / 2q² + 1 / 2s²)².
Using the identity (A - B)² - (A + B)² = -4AB, we simplify:
d² - (R₁ + R₂)² = (ps - qr)² / (q²s²) - 1 / (q²s²) = ((ps - qr)² - 1) / (q²s²).
Since p / q and r / s are distinct fractions, ps ≠ qr. Because p, q, r, and s are integers, |ps - qr| ≥ 1.
If |ps - qr| = 1, the numerator is 0, so d² = (R₁ + R₂)². The distance equals the sum of the radii, so the circles are exactly tangent.
If |ps - qr| > 1, then (ps - qr)² - 1 > 0, so d² > (R₁ + R₂)². The distance is strictly greater than the sum of the radii, so the circles are wholly external. They can never intersect in two points.
Original problem 105
Show that the representative Ford circles of two distinct fractions are tangent at one point precisely when the fractions are adjacent in some Farey sequence Fₙ.
Proof of problem 105
From the proof in Exercise 104, two Ford circles C(p, q) and C(r, s) are tangent if and only if:
(ps - qr)² - 1 = 0, so |ps - qr| = 1.
The condition |ps - qr| = 1 is the fundamental unimodular property that defines two fractions being adjacent in a Farey sequence.
Specifically, if |ps - qr| = 1, the fractions p / q and r / s will be consecutive terms in the Farey sequence Fₙ where n = max(q, s).
Therefore, tangency occurs if and only if the fractions are adjacent in some Farey sequence.
area(R) > 4
Translating pieces of a region into one unit square forces overlap once total area exceeds 1. Scaling a centrally symmetric convex region by one half, then using overlap, symmetry, and convexity, produces a nonzero lattice point in the original region.
Original problem 121
Show that there must exist i, j and m, n such that Tᵢⱼ ∩ Tₘₙ ≠ ∅, and i ≠ m or j ≠ n.
Proof of problem 121
By definition, the sets Rᵢⱼ are mutually disjoint, since they are intersections of R with disjoint grid squares Iᵢⱼ, and their union exactly constitutes the region R.
Thus, the sum of their areas equals the area of R:
Σᵢⱼ A(Rᵢⱼ) = A(R).
We are given that A(R) > 1.
The set Tᵢⱼ is created by translating Rᵢⱼ by the integer vector -(i, j). Because translation strictly preserves area, A(Tᵢⱼ) = A(Rᵢⱼ).
Furthermore, the translation shifts every piece into the unit square S = [0, 1) × [0, 1), meaning Tᵢⱼ ⊂ S for all i, j.
Assume for the sake of contradiction that all the translated pieces Tᵢⱼ are mutually disjoint. Then the total area of their union would be the sum of their individual areas, and this union would be entirely contained within S:
Area(⋃ᵢⱼ Tᵢⱼ) = Σᵢⱼ A(Tᵢⱼ) ≤ A(S) = 1.
However, Σᵢⱼ A(Tᵢⱼ) = Σᵢⱼ A(Rᵢⱼ) = A(R) > 1. This creates a contradiction.
Therefore, our assumption is false, and at least two translated sets must overlap. There must exist (i, j) ≠ (m, n) such that Tᵢⱼ ∩ Tₘₙ ≠ ∅.
Original problem 122
Complete the proof of Blichfeldt's Theorem.
Proof of problem 122
From Exercise 121, there exists an overlapping point p ∈ Tᵢⱼ ∩ Tₘₙ with (i, j) ≠ (m, n).
Because p ∈ Tᵢⱼ, it was translated from some point p₁ = (x₁, y₁) ∈ Rᵢⱼ ⊂ R. Thus, p₁ - (i, j) = p.
Because p ∈ Tₘₙ, it was also translated from some point p₂ = (x₂, y₂) ∈ Rₘₙ ⊂ R. Thus, p₂ - (m, n) = p.
Equating these expressions gives p₁ - (i, j) = p₂ - (m, n).
Rearranging gives:
p₁ - p₂ = (i, j) - (m, n) = (i - m, j - n).
Since i, j, m, and n are integers, the difference vector (i - m, j - n) consists entirely of integers. Therefore, x₁ - x₂ ∈ Z and y₁ - y₂ ∈ Z.
Finally, since we chose distinct grid intervals, the points p₁ and p₂ must be distinct. This finds two distinct points in R whose difference is an integer point, completing the proof of Blichfeldt's Theorem.
Original problem 124
Let R be a bounded, convex region in R² that is symmetric about the origin and has area greater than 4. Consider R′ = {(1 / 2)x such that x ∈ R}. Since R′ is a smaller version of R, it is convex and symmetric about the origin. Show that there are points x′ and y′ in R′ such that x′ - y′ is a nonzero lattice point.
Proof of problem 124
Blichfeldt's Theorem states that for any measurable set S ⊂ Rⁿ with volume greater than 1, there exist two distinct points u, v ∈ S such that u - v ∈ Zⁿ \ {0}.
First, calculate the area of R′. Since R′ is defined by scaling the coordinates of R by a factor of 1 / 2, the area scales by the square of that factor:
Area(R′) = (1 / 2)² Area(R) = (1 / 4)Area(R).
Given that Area(R) > 4, we have Area(R′) > (1 / 4)(4) = 1.
Since Area(R′) > 1, Blichfeldt's Theorem gives distinct points x′, y′ ∈ R′ such that x′ - y′ ∈ Z² \ {0}.
This confirms there is a nonzero lattice point formed by the difference of two points in R′.
Original problem 125
Let x′ and y′ be as in Exercise 124. Show that x′ - y′ is in R. Hint: express x′ - y′ as a linear combination of points that you know are in R.
Proof of problem 125
By the definition of R′, since x′, y′ ∈ R′, there exist points x, y ∈ R such that x′ = (1 / 2)x and y′ = (1 / 2)y.
We examine the difference:
x′ - y′ = (1 / 2)x - (1 / 2)y = (1 / 2)x + (1 / 2)(-y).
Since R is symmetric about the origin, if y ∈ R, then -y ∈ R.
Since R is convex, any convex combination of points in R must also be in R.
The expression (1 / 2)x + (1 / 2)(-y) is a convex combination, specifically the midpoint, of x and -y, both of which are in R.
Therefore, x′ - y′ ∈ R. This proves that there exists a nonzero lattice point inside R, which concludes the proof of Minkowski's Theorem.
6 / π²
A lattice point is visible from the origin exactly when its coordinates are coprime. Möbius inversion separates points by greatest common divisor, and the reciprocal zeta identity at 2 turns the limiting proportion into 6 / π².
Original problem 16
The distribution of visible lattice points in the plane is 6 / π². Consider the square region S(t) in the plane defined by |x| ≤ t and |y| ≤ t. Let N(t) denote the number of lattice points in this square, and let V(t) denote the number of lattice points in the square that are visible from the origin. Show that limt→∞ V(t) / N(t) = 6 / π².
Proof
A lattice point (x, y) is visible from the origin if and only if gcd(x, y) = 1.
The total number of lattice points N(t) in [-t, t] × [-t, t] is (2⌊t⌋ + 1)². As t → ∞, N(t) ≈ 4t².
To find V(t), sum over all possible values of d = gcd(x, y). If gcd(x, y) = d, then x = da and y = db, where gcd(a, b) = 1.
Using the property of the Möbius function μ(d), where Σd | gcd(x,y) μ(d) equals 1 if gcd(x, y) = 1 and 0 otherwise, we can count visible points:
V(t) = Σ|x|≤t Σ|y|≤t Σd | gcd(x,y) μ(d).
Rearranging the sums and letting x = da and y = db gives:
V(t) = Σd=1⌊t⌋ μ(d) · (number of lattice points in [-t/d, t/d] × [-t/d, t/d]).
The number of points in the scaled square is (2⌊t / d⌋ + 1)² ≈ 4t² / d².
Therefore:
V(t) ≈ Σd=1∞ μ(d)(4t² / d²) = 4t² Σd=1∞ μ(d) / d².
Recalling that Σn=1∞ μ(n) / nˢ = 1 / ζ(s), for s = 2 we have:
Σd=1∞ μ(d) / d² = 1 / ζ(2) = 1 / (π² / 6) = 6 / π².
Thus V(t) ≈ 4t² · 6 / π². Taking the limit of the ratio gives:
limt→∞ V(t) / N(t) = [4t²(6 / π²)] / 4t² = 6 / π².
Stanford Online High School
Graduated with a 3.9/4.0 unweighted GPA and a 1550 SAT score.
Translator on yachts at Club Náutico de Águilas throughout high school.
I translated between English, Spanish, and Polish for clients, crews, and staff.
I also translated all the articles for my dad's personal journal into Spanish.
I grew up speaking Polish, Spanish, and Czech, and travelling around Europe. I also speak Mandarin Chinese, and I am hoping to add French to this list.
Polish
Spanish
Czech
Mandarin Chinese
Through high school, my plan was mathematics at the University of Edinburgh: an hour from home, in a city I loved that was deeply intertwined with my life and family.
I worked tirelessly in school towards that future. Then my school counsellor submitted my applications five days late, lied to me about it and my offer was rescinded.
It crushed the future I had dreamt of and worked towards for four years. I spiralled into a major existential crisis, because for the first time I didn't know what I wanted to do with my life. I became completely disillusioned with the college admission process, because why work so hard just to put your future in the hands of people who don't know you or even care to know you?
I was very low during this period of my life. I had teachers judge and ignore me at my own graduation because I didn't get into college. I witnessed parents of friends shun my mother for it. I had to stand in front of a podium at Stanford in front of hundreds of people and their families and talk about how much hope they had in their future, despite not knowing what awaited in my own.
A lot of things during this time pointed me towards entrepreneurship, though I was too down in the dumps to see it. During a graduation party I met a photographer who turned out to own a really big startup in SF, and we talked for hours about AI and how to travel the world when you're broke. He told me that taking a break from academics would be the greatest thing to ever happen to me, and that I should take that time to really think about what it is I want from this life.
I was utterly bored over the summer so I took a train to Newcastle where I went to an open day at Durham on my own. I sat through a few lectures, talked to a lot of students there, and found myself utterly disappointed. I found out, to my dismay, that the curriculums even at the "best" universities here in the UK wouldn't cover any math that I didn't already know. Nor would they cover any new philosophy, literature, or anything I was interested in.
I got the most backhanded things said to me by the head of the philosophy department for asking too many questions, and all the undergrads I talked to asked me what my PhD was in, which was very funny to me because usually I get mistaken for someone much younger. I felt ostracized and lost.
That was until I wandered into this little booth run by one guy. I asked him what he was advertising, and he told me a bit about the venture society at Durham. I said, "Oh wow that's cool, the only thing that can save the UK from its horrible economic policies at this point is more entrepreneurship," because I'd read it in an article somewhere, and he lit up. We talked for ages about entrepreneurship and UK culture and how much we both wished we lived in SF.
This was the only place in the whole university where I felt I belonged, in the same way the only person who seemed genuinely excited for my future at that graduation party was Mr. photographer-entrepreneur. Somehow, I found that I no longer belonged in academia. Or maybe I never belonged! Who knows. Either way, I figured I'd stick around these startup people because they're the only people I really like being around.
A month later, my best friend and I had a wild idea to circumvent social media bans. This marks my first ever startup. Working on this project, and actually monetizing it, has been one of the most exciting and fulfilling things I have done all year. I want nothing else in life than to work on crazy projects with cool people.
So no, I will no longer place my life in the hands of admissions officers and counsellors who do not care to know me. I will build a life around work I genuinely believe in. Very excited for what comes next!