It turns out that 12 cm is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way, but there are many more examples.

12 cm: (3,4,5)
24 cm: (6,8,10)
30 cm: (5,12,13)
36 cm: (9,12,15)
40 cm: (8,15,17)
48 cm: (12,16,20)

In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths allow more than one solution to be found; for example, using 120 cm it is possible to form exactly three different integer sided right angle triangles.

120 cm: (30,40,50), (20,48,52), (24,45,51)

Given that L is the length of the wire, for how many values of L ≤ 2,000,000 can exactly one integer sided right angle triangle be formed?

Analysis

This is an extension to the Problem 9 solution.

Solution

Runs < 3 seconds in Python.

from Euler import gcd
 
limit = 2000000
sqrt_limit = int(limit**.5)
tx = [0]*limit
for i in xrange(1, sqrt_limit, 2):
  for j in xrange(2, sqrt_limit, 2):
    if gcd(i, j) == 1: 
      sum = abs(j*j - i*i) + 2*i*j + i*i + j*j
      for s in range(sum, limit, sum):
        tx[s]+=1
 
print "Answer to PE75 = ", tx.count(1)

Comments

If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.

{20,48,52}, {24,45,51}, {30,40,50}

For which value of p ≤ 1000, is the number of solutions maximised?

Analysis

The Pythagorean Theorem gives us: a² + b² = c² and by intuition a + b + c = p (perimeter).
The decision to iterate perimeters has many advantages over generating right triangles. Only even numbered perimeters need to be checked; below is our analysis:

No matter which integral values we choose for a, b and c such that a² + b² = c² the perimeter will be even.

By iterating perimeters we need a method to check for integral (integer) right triangles.
We can interpret the Pythagorean theorem as a² + b² = (p-a-b)² since c = p-a-b. After expanding and simplifying we have:
b = p(p – 2a) / 2(p-a) and when this equation evaluates to an integer (MOD(p(p – 2a), 2(p-a)) = 0) we have our triangle.

One last optimization. By using a+b>c and symmetry we only need to investigate values of a to p/4.

Solution

Runs < 1 second in Python.

t_max = 0;
p_limit = 1000
 
for p in xrange (2, p_limit+1, 2):
  t = 0;
  for a in xrange (2, p/4+1):
    if  p*(p - 2*a) % (2*(p-a)) == 0: t += 1
    if t > t_max: (t_max, p_max) = (t, p)
 
print "Answer to PE30 = ", p_max

Comments

Some other results:
10,000 = 9,240
100,000 = 55,440
1,000,000 = 720,720
10,000,000 = 6,126,120