100 số pell đầu tiên

Pell numbers carry the wrong name, and the slip is Euler's. The equation they solve, x² − 2y² = ±1, belongs to a family that came to English attention in 1657, when Pierre de Fermat challenged mathematicians there to solve x² − Ny² = 1 for arbitrary N. William Brouncker, shortly afterwards the first president of the Royal Society, produced a working method, and John Wallis published the exchange. Reading it decades later, Leonhard Euler credited the method to John Pell (1611–1685) instead, and "Pell's equation" has been the name ever since. Euler may not have been simply careless: the equation does appear in Johann Rahn's algebra of 1659, a book written with Pell's help and, by some accounts, largely written by Pell — so the misattribution is arguable rather than plainly wrong.

The substantive history is older and further east. Brahmagupta, writing in 628, gave a composition rule for combining solutions and used it to solve x² − 92y² = 1, finding x = 1151, y = 120. By the twelfth century Bhāskara II had the cyclic chakravala method, which handles any N; a surviving fragment attributed to Jayadeva suggests it was known a century or two earlier. Lagrange supplied the first proof, in the 1760s, that a solution always exists when N is not a perfect square.

The numbers themselves are older still, because N = 2 is simply the problem of approximating √2. Theon of Smyrna, in the second century, described what he called side and diameter numbers: begin with side 1 and diameter 1, then replace them repeatedly by side + diameter and twice the side plus the diameter. The sides that emerge are 1, 2, 5, 12, 29, 70 — the Pell numbers — and the ratios 1/1, 3/2, 7/5, 17/12, 41/29, 99/70 are exactly the continued-fraction convergents of √2.

Tính chất chính

• P(0) = 0, P(1) = 1, and P(n) = 2·P(n−1) + P(n−2).
• P(n+1)/P(n) converges to the silver ratio 1 + √2 ≈ 2.4142135624, the positive root of x² = 2x + 1.
• With the half-companion numbers H = 1, 1, 3, 7, 17, 41, 99, the identity H(n)² − 2·P(n)² = (−1)ⁿ holds exactly, so each pair solves the Pell equation for N = 2.
• H(n)/P(n) are precisely the continued-fraction convergents of √2, since √2 = [1; 2, 2, 2, …], and they fall alternately below and above it.
• P(n)² + P(n+1)² = P(2n+1): the sum of two consecutive squares in this sequence is another Pell number.
• The kth square triangular number is (P(2k)/2)² — giving 1, 36, 1225, 41616 and 1413721, which are the triangular numbers of index 1, 8, 49, 288 and 1681.
• If m divides n then P(m) divides P(n); more strongly, gcd(P(m), P(n)) = P(gcd(m, n)).
• The only Pell numbers that are perfect squares — or any higher perfect power — are 0, 1 and 169 = 13².