# Find the number of lengths of wire below 2 000 000
# that can form a right-angled triangle in exactly one way
from useful import gcd
oneWay = {}
n = 1
m = 2
while m*m + 2*m*n + m*m <= 2000000 :
val = 2*m*(m+n)
while val <= 2000000 :
if gcd(m,n) == 1 :
count = 1
while count*val <= 2000000 :
if count*val in oneWay :
oneWay[count*val] = False
else :
oneWay[count*val] = True
count += 1
m += 2
val = 2*m*(m+n)
n += 1
m = n+1
print len([key for key in oneWay.keys() if oneWay[key] == True])
max_N = 0
for N in xrange(1, 1001) :
root = sqrt(N)
floor_root = int(root)
if floor_root * floor_root == N :
# Perfect square
continue
left_of_frac, frac_top, frac_bottom = floor_root, 1, floor_root
lis = []
print N
while True :
lis.append(left_of_frac)
new_frac_bottom_before_reduce = N - frac_bottom*frac_bottom
divisor = gcd(frac_top, new_frac_bottom_before_reduce)
new_frac_top_int_part = new_frac_bottom_before_reduce / divisor
assert(frac_top == divisor)
new_frac_top_before_reduce = (sqrt(N) + frac_bottom)
new_left_of_frac = int(new_frac_top_before_reduce / new_frac_top_int_part)
new_frac_top = (new_frac_top_before_reduce -
new_left_of_frac * (new_frac_top_int_part))
new_frac_bottom = int(sqrt(N) - new_frac_top + 0.0001)
left_of_frac, frac_top, frac_bottom = (new_left_of_frac,
new_frac_top_int_part, new_frac_bottom)
# Update the convergent fraction
x, y = 1, 0
for left_term in reversed(lis) :
y, x = x, y
x, y = left_term * y + 1 * x, y
def fracreduce((p1, q1)) : d = gcd(p1,q1) return (p1/d, q1/d)
# In a sorted list of all reduced proper fractions where the denominator <= 1000000, # find the fraction to the left of 3/7 from useful import gcd lowest = (2,7) for denom in range(2,1000001) : if denom == 7: continue x = 3 * denom / 7 while (x+1)*7 < 3*denom : x += 1 if gcd(x,denom) != 1 : continue if lowest[1] * x > lowest[0] * denom : lowest = (x,denom) print lowest
# This program does not terminate in a timely fashion, because we use a
# very generous upper bound for m, but it does output the solution within
# 1 minute
from math import sqrt
from useful import gcd
top = 100 * 1000 * 1000
count = 0
for m in xrange(top) :
if m % 2 == 1 :
lower = 2
else :
lower = 1
for n in xrange(lower, m, 2) :
if gcd(m, n) != 1 :
continue
x = m*m - n*n
y = 2*m*n
z = m*m + n*n
if z % (x-y) == 0 :
# For what scaling values of k is k * (x + y + z) < 100 million?
k = top / (x+y+z)
count += k
print (x,y,z), k, count
print count