def main():
# for i in range(2, 10000):
# if is_square(5 * i**2 + 14 * i + 1):
# print(i)
# print(i * 5 + 7, math.sqrt(((i * 5 + 7)**2 - 44) // 5))
# pell equation u**2 - 5 * v**2 = 44
# fundamental solutions found through trial
fund_uv_s = [(17, 7), (32, 14), (112, 50),
(217, 97), (767, 343), (1487, 665)]
fund_x, fund_y = pell.pell(5)
nuggets = []
for u, v in fund_uv_s:
for _ in range(N):
if u % 5 == 2:
nuggets.append((u - 7) // 5)
# print((u - 7) // 5, total)
u, v = u * fund_x + 5 * v * fund_y, v * fund_x + u * fund_y
# print(u, v)
# print((u - 7) / 5)
nuggets.sort()
# print(nuggets)
print(sum(nuggets[:N]))
def main():
# for s in range(2, 10000):
# sol = rational_quad_solve(s, s + 1, -s)
# if sol is not None:
# x = list(filter(lambda x: x > 0, sol))[0]
# # print(s, x)
# # print(s + 1, 2 * s, int(math.sqrt((s + 1)**2 + 4 * s * s)))
# u = s * 5 + 1
# print(u, (u ** 2 + 3) * u // 2)
# Find Solution
# u**2 - 5 * v**2 = -4
# x = (u**2+3) * u / 2
# y = (u**2+1) * v / 2
# fundamental solution to pell equation
fund_x, fund_y = pell.pell(5, True)
x, y = fund_x, fund_y
# solution exists for every 2
for i in range(N * 2):
x, y = pell.next_solution(5, fund_x, fund_y, x, y, True)
# solve equation of the form 2*x = u**3 + 3 * u
# intuition: 2*x is close to 2*u**3
# because of floating point inprecision,
# starting from (2*x)**(1/3) is closer
for u in range(int((2 * x)**(1 / 3)), int((2 * x)**(1 / 3)) * 2):
# print((u**2 + 3) * u - x * 2)
if (u**2 + 3) * u == x * 2:
break
# print(x, y, u)
print((u - 1) // 5)
def main():
# for s in range(2, 10000):
# sol = rational_quad_solve(s, s + 1, -s)
# if sol is not None:
# x = list(filter(lambda x: x > 0, sol))[0]
# # print(s, x)
# # print(s + 1, 2 * s, int(math.sqrt((s + 1)**2 + 4 * s * s)))
# u = s * 5 + 1
# print(u, (u ** 2 + 3) * u // 2)
# Find Solution
# u**2 - 5 * v**2 = -4
# x = (u**2+3) * u / 2
# y = (u**2+1) * v / 2
# fundamental solution to pell equation
fund_x, fund_y = pell.pell(5, True)
x, y = fund_x, fund_y
# solution exists for every 2
for i in range(N * 2):
x, y = pell.next_solution(5, fund_x, fund_y, x, y, True)
# solve equation of the form 2*x = u**3 + 3 * u
# intuition: 2*x is close to 2*u**3
# because of floating point inprecision,
# starting from (2*x)**(1/3) is closer
for u in range(int((2 * x)**(1 / 3)), int((2 * x)**(1 / 3)) * 2):
# print((u**2 + 3) * u - x * 2)
if (u**2 + 3) * u == x * 2:
break
# print(x, y, u)
print((u - 1) // 5)
def main():
result = 0
for a, b in pell.pell(3):
if a < 7:
continue
if 2*a+2 > 1000000000:
break
if (a+2) % 3 == 0:
result += 2*a+2
if 2*a-2 > 1000000000:
break
if (a-2) % 3 == 0:
result += 2*a-2
print result
def main():
areas_l = set()
# fundamental solution to x^2 - 5y^2 = 1
x, y = pell.pell(5)
first_x, first_y = x, y
areas_l.add(get_area_l(x, y))
for _ in range(N):
x, y = pell.next_solution(5, first_x, first_y, x, y)
areas_l.add(get_area_l(x, y))
# fundamental solution to x^2 - 5y^2 = -1
x, y = pell.pell(5, True)
first_x, first_y = x, y
areas_l.add(get_area_l(x, y))
for _ in range(N):
x, y = pell.next_solution(5, first_x, first_y, x, y)
x, y = pell.next_solution(5, first_x, first_y, x, y)
areas_l.add(get_area_l(x, y))
areas_l = sorted(areas_l)
print(sum(areas_l[i][1] for i in range(N)))
def main():
areas_l = set()
# fundamental solution to x^2 - 5y^2 = 1
x, y = pell.pell(5)
first_x, first_y = x, y
areas_l.add(get_area_l(x, y))
for _ in range(N):
x, y = pell.next_solution(5, first_x, first_y, x, y)
areas_l.add(get_area_l(x, y))
# fundamental solution to x^2 - 5y^2 = -1
x, y = pell.pell(5, True)
first_x, first_y = x, y
areas_l.add(get_area_l(x, y))
for _ in range(N):
x, y = pell.next_solution(5, first_x, first_y, x, y)
x, y = pell.next_solution(5, first_x, first_y, x, y)
areas_l.add(get_area_l(x, y))
areas_l = sorted(areas_l)
print(sum(areas_l[i][1] for i in range(N)))
def main():
threshold = 10 ** 12
fund_x, fund_y = pell.pell(2, True)
prev_x, prev_y = fund_x, fund_y
m = (prev_x + 1) // 2
n = (prev_y + 1) // 2
while m < threshold:
prev_x, prev_y = pell.next_solution(2, fund_x, fund_y, prev_x, prev_y, True)
m = (prev_x + 1) // 2
n = (prev_y + 1) // 2
# print(m-n, n)
m = (prev_x + 1) // 2
n = (prev_y + 1) // 2
print(n)
def main():
threshold = 10**12
fund_x, fund_y = pell.pell(2, True)
prev_x, prev_y = fund_x, fund_y
m = (prev_x + 1) // 2
n = (prev_y + 1) // 2
while m < threshold:
prev_x, prev_y = \
pell.next_solution(2, fund_x, fund_y, prev_x, prev_y, True)
m = (prev_x + 1) // 2
n = (prev_y + 1) // 2
# print(m-n, n)
m = (prev_x + 1) // 2
n = (prev_y + 1) // 2
print(n)
def test_pell50(self):
x = pell(49)
y = pell(50)
self.assertEqual(x, 2015874949414289041)
self.assertEqual(y, 4866752642924153522)
'''
Created on 2011-7-13
@author: huangkan
'''
import pell
max=3
ans=2
for i in range(3,1001):
if i**.5!=int(i**.5):
t,t2=pell.pell(i)
if t>max:
max=t
ans=i
print(ans)
