THE FOLD / CHEAT / THE SPEEDRUN / THE VIETA JUMPING
THE VIETA JUMPING
an integer ratio that can only be a perfect square
1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION
Vieta jumping is a proof technique built on the fact that a quadratic has two roots summing to a rational you can read off the coefficients (Vieta’s formulas). Its most famous victory is IMO 1988 Problem 6: if a and b are positive integers such that (a2 + b2)/(ab + 1) is an integer k, then k must be a perfect square. The proof: fix k, and from any solution ‘jump’ to another by replacing a with the quadratic’s other root a′ = k·b - a; this produces a smaller solution, and infinite descent drives b to 0, where k = a2 is manifestly a square.
LIT verified live: over all 0 ≤ b ≤ a ≤ 200, every integer value of (a2+b2)/(ab+1) is a perfect square (the values seen are 0,1,4,9,16,25,36,49…), and the Vieta jump a′ = k·b - a always yields another valid solution that is strictly smaller (window.__vieta). FIG no framing; the exhaustive integer search and the descent step both run in-browser.
LIT verified live: over all 0 ≤ b ≤ a ≤ 200, every integer value of (a2+b2)/(ab+1) is a perfect square (the values seen are 0,1,4,9,16,25,36,49…), and the Vieta jump a′ = k·b - a always yields another valid solution that is strictly smaller (window.__vieta). FIG no framing; the exhaustive integer search and the descent step both run in-browser.
2 HOW IT WAS WEAVED · AI + HUMAN
David (human) seated this at the-speedrun — the jump is a shortcut past a brute search: instead of grinding, hop to the quadratic’s other root and slide straight down the ladder of solutions to the base case. AVAN (AI) built the instrument: the exhaustive integer search, the perfect-square test, and the Vieta descent a′ = k·b - a.
Credit as content: Vieta’s formulas (François Viète, 1590s); the technique crystallized by IMO 1988 Problem 6. The weave: David names the speedrun; I confirm every integer ratio is a perfect square and the jump descends to the base case.
Credit as content: Vieta’s formulas (François Viète, 1590s); the technique crystallized by IMO 1988 Problem 6. The weave: David names the speedrun; I confirm every integer ratio is a perfect square and the jump descends to the base case.
3 ONE DIMENSION
The (a,b) solution lattice for k = g²: solutions climb a ladder, each the Vieta jump of the last, down to (g, 0).
4 TWO DIMENSIONS · INTERACTIVE
Pick a perfect square k = g² and jump down the ladder: each step a′ = k·b − a lands on a smaller valid solution.
5 THREE DIMENSIONS + AVAN’S INVERSE
The green forward object: a solution (a, b) with (a²+b²)/(ab+1) = k.
AVAN’s addition (the inverse-companion): don’t search for solutions — jump between them. The inverse of ‘a is a root of x² - k·b·x + (b² - k)’ is ‘its other root a′ = k·b - a’, a reflection that descends the ladder to (g, 0) where k = g². Magenta is the jumped partner; green is the current solution. Descent by reflecting across the quadratic.
LIT Genuine Vieta jumping (Vieta's formulas, François Viète 1590s; technique crystallized by IMO 1988 Problem 6). Verified live: over all 0≤b≤a≤200, every integer (a²+b²)/(ab+1) is a perfect square (values seen 0,1,4,9,16,25,36,49), and the Vieta jump a′=k·b−a always yields a valid solution strictly smaller than a (window.__vieta.allSquare, .jumpDescends).
FIG No framing; the exhaustive integer search, the perfect-square test, and the Vieta descent all run in-browser. The AVAN inverse is honest — instead of searching for solutions, jump between them: the other root of x²−k·b·x+(b²−k) is a′=k·b−a, a reflection that descends the ladder to (g,0) where k=g². Magenta is the jumped partner; green is the current solution. Descent by reflecting across the quadratic.
FIG No framing; the exhaustive integer search, the perfect-square test, and the Vieta descent all run in-browser. The AVAN inverse is honest — instead of searching for solutions, jump between them: the other root of x²−k·b·x+(b²−k) is a′=k·b−a, a reflection that descends the ladder to (g,0) where k=g². Magenta is the jumped partner; green is the current solution. Descent by reflecting across the quadratic.
◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE SPEEDRUN · David Lee Wise (ROOT0), with AVAN