'''
This is just an extension of Problem 9 of project euler.
I just initiated the value, then took a squareroot( since the values req. are 3)
Then did a gcd b/w the 2 values to find the later value.
Then did a sum to get the Length. Then added them and printed them out.
'''
from Euler import gcd, sqrt
L = 1500000
sqrt_L = int(sqrt(L))
lengths = [0]*L #initating the arrays.
for i in range(1, sqrt_L, 2):
for j in range(2, sqrt_L- i , 2):
if gcd(i, j) == 1:
sum = abs(j*j - i*i) + 2*i*j + i*i + j*j
for s in range(sum, L, sum):
lengths[s]+=1
print lengths.count(1)
Assume there are no solutions in positive integers when D is square.
By finding minimal solutions in x for D = {2, 3, 5...}, we obtain the following:
3^2 – 2×2^2 = 1
2^2 – 3×1^2 = 1
> 9^2 – 5×4^2 = 1
5^2 – 6×2^2 = 1
8^2 – 7×3^2 = 1
Hence, for D ≤ 6, the largest x is obtained when D=5.
Find the value of D ≤ 1000 in minimal solutions of x
for which the largest value of x is obtained.
'''
from Euler import sqrt, cfrac, convergents
def Pell(D):
chain = cfrac(D)
if len(chain)==1: return None
for f in convergents(chain):
x = f.numerator
y = f.denominator
if x**2 - D*y**2 == 1:
return x
ans = max([D for D in range(1000) if not sqrt(D).is_integer()], key=Pell)
print(ans)
from Euler import prime_sieve, sqrt
L = 10**7
primes = prime_sieve(int(1.30 * sqrt(L)))
del primes[:int(0.6 * len(primes))]
def Eluer70(limit):
min_q, min_n, i = 2, 0, 0
for p1 in primes:
i += 1
for p2 in primes[i:]:
if (p1 + p2) % 9 != 1: continue
n = p1 * p2
if n > limit: return min_n
phi = (p1 - 1) * (p2 - 1)
q = n / float(phi)
if sorted(str(phi)) == sorted(str(n)):
if min_q > q:
min_q = q
min_n = n
print Euler70(L)
"""
This is just an extension of Problem 9 of project euler.
I just initiated the value, then took a squareroot( since the values req. are 3)
Then did a gcd b/w the 2 values to find the later value.
Then did a sum to get the Length. Then added them and printed them out.
"""
from Euler import gcd, sqrt
L = 1500000
sqrt_L = int(sqrt(L))
lengths = [0] * L # initating the arrays.
for i in range(1, sqrt_L, 2):
for j in range(2, sqrt_L - i, 2):
if gcd(i, j) == 1:
sum = abs(j * j - i * i) + 2 * i * j + i * i + j * j
for s in range(sum, L, sum):
lengths[s] += 1
print lengths.count(1)
from Euler import prime_sieve, is_perm, sqrt
min_q, i, L = 2, 0, 10**7
primes = prime_sieve(1.30*sqrt(L))
ll = 0.7*sqrt(L)
for n in range(len(primes)):
if primes[n]>ll: break
del primes[:n]
for p1 in primes:
i+=1
for p2 in primes[i:]:
n = p1 * p2
if n > L: break
phi = (p1-1) * (p2-1)
q = n / float(phi)
if is_perm(phi, n) and min_q>q: min_q, min_n = q, n
print "Answer to PE70 = ",min_n
