A 1000 Pythagorean triplets

Today’s problem in Project Euler, is related to probably THE most well known theorem of all times. Pythagoras theorem stating that

The square of the hypotenuse is the sum of squares of the two other sides.

It can also be stated in a more geometrical way as

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares
whose sides are the two legs (the two sides that meet at a right angle

Which can also be shown in a graphical sense, on the figure to the right, where the blue area equals the orange area.

But enough with the background info, the problem reads

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a² + b² = c²

For example, 3² + 4² = 9 + 16 = 25 = 5².

There exists exactly one Pythagorean triplet for which a + b + c = 1000.

Find the product abc.

There is an infinite number of Pythagorean triplets, so I am quite glad for the last condition stating that the circumference of the triangle has to be 1000. As a side note Fermat conjectured, and wrote that he had a proof in the margin of his text book that aⁿ + bⁿ = cⁿ has no integer solutions for n > 2. It has only recently been proven that he was right.

A great book on that topic is Fermat’s Last Theorem By Simon Singh. But that is just a side track.

There are two approaches for finding a solution in my post, there is the brute force method, which is widely described on the net, and works brilliantly for small numbers, and then there is a number theoretical approach which I myself had to study. So providing the latter is a test for me to see if I understood it to a level where I can explain it to you. And don’t hesitate to ask for clarification or point out errors in my deductions.

In my further deduction I will denote the circumference of the triangle s, which in the problem is fixed to 1000.

Brute Force

For the brute force algorithm we would need to loop over all the possible values of a and b, calculatec and check if that is a solution. We could be ignorant and just let a and b loop from 1-1000, or we could use the facts that a < b < c, and thus exploit that  a < s/3,  and a < b < s/2.

The code can be written as

int a = 0, b = 0, c = 0;
int s = 1000;
bool found = false;
for (a = 1; a < s / 3; a++) {
    for (b = a; b < s / 2; b++) {
        c = s - a - b;
        if (a * a + b * b == c * c) {
            found = true;
            break;
        }
    }
 
    if (found) {
        break;
    }
}

 

and the result

The Pythagorean triple is 200, 375, 425, and the sum is 1000
The product is 31875000
Solution took 0 ms

Number Theoretical Approach

As I mentioned in the beginning of the post, this approach took me a while to figure out, so I have made my explanation rather elaborate to make it understandable for myself.  The rest of this post will be more mathematical rigorous than any of my previous posts. I hope you enjoy reading it as much as I did writing it.

Prerequisites and Definitions

More than 2000 years ago Euclid established that Pythagorean Triplets can be generated by the formula.

Given any positive integers m and n where m > n > 0, the integers

a =m²-n²

b =2m*n

c =m²+n²

forms a Pythagorean triplet.  You can convince your self that (m²-n²)² + (2m*n)² = (m²+n²)² by expanding both sides.  There are quite a number of other formulas for generating triplets, but this is the one I will use here. Depending on the choices of m and n, we might have to interchange a and bto adhere to the constraint a < b.

Furthermore we need a definition of primitive.

Definition: Primitive

A Pythagorean Triplet is called primitive if a, b and c are coprime. That means that the greatest common divisor (gcd) of the set {a,b,c} is one.

A Pythagorean triplet is primitive if and only if  m and n are coprimes and exactly one of them is even.

Proof

Showing that it is a necessary condition is rather easy. Assume that gcd(n,m)= d > 1, then d² is a common divisor for {a,b,c} (you can convince your self about it by inserting m = d*p and n = d*q), and if both n and m are odd then a,b and c are even, and thus 2 is a common divisor. Thus these conditions are necessary.

Showing that they are sufficient is a bit more tricky.

Assume that exactly one of  m and n is even and and triplet is not primitive. Since the triplet is not primitive a prime p is divisible with a, b and c (according to the fundamental theorem of arithmetic as referred to in  problem 3 such a prime exists).

From the condition that exactly one of m and n is odd, it follows that a and c is also odd, and so is pthen. Any odd prime that that divides b must also divide m and/or n, without loss of generality assume that it divides m. Thus it must also divide m². Since the prime is also a divisor of c = m² + n², then it follows that it must divide n² and n as well. (this follows from notions and facts part of the divisor article on wikipedia) Therefore p is a divisor of m and n, and therefore m and n are not coprimes. Thus it is sufficient that the conditions are fulfilled.

Finding a new unique representation

Every primitive Pythagorean triplet can be represented by a unique pair of coprimes m and n where exactly one of them is even. However, not all Pythagorean triplets are generated this way. Approaching it from the other end. From every Pythagorean triplet a primitive triplet can be generated by dividing by the greatest common divisor, thus every Pythagorean triplet can be represented by the unique set {d,m,n} by the given formula

a =d(m²-n²)

b =2d*m*n

c =d(m²+n²)

with m > n > 0, m and n being coprimes and exactly one of m and n even. d is the greatest common divisor of a, b and c.

Solving the problem

Now we have a new unique representation of any triplet, but how does that help us solving the problem?

Using the conditions

a + b+ c = s = d(m²-n²) + 2d*m*n + d(m²+n²) = 2m(m+n)d

We can see that m must be a divisor to s/2, and

and we need to find a k = m+n such that k is a divisor to s/(2m) and

m <k < 2m,

k < s/(2m),

k is odd and gcd(m,k) = 1.

In our problem it means that m < 23, thus decreasing the first number we need to find by a factor 10. The second number k, we need to find has an upper bound depending on m but is in the same vicinity. Once we have found m and k, the rest can be calculated easily. So if we can find a cheap way to calculate the greatest common divisor, we can make an efficient algorithm for finding the triplet.

Algorithm for greatest common divisor

On wikipedia there is a few different methods for calculating the greatest common divisor. I have chosen Euclid’s  algorithm, which is fairly fast and easy to implement. The code can be written as

public int gcd(int a, int b) {
    int y = 0;
    int x = 0;
 
    if (a > b) {
        x = a;
        y = b;
    } else {
        x = b;
        y = a;
    }
 
    while (x % y != 0) {
        int temp = x;
        x = y;
        y = temp % x;
    }
    return y;
}

 

Finding the Pythagorean triplet

We now have all parts we need to construct an algorithm, and a straight forward approach is

int a=0, b=0, c=0;
int s = 1000;
int m = 0, k = 0, n = 0, d = 0;
bool found = false;
 
int mlimit = (int) Math.sqrt(s / 2);
for (m = 2; m <= mlimit; m++) {
    if ((s / 2) % m == 0) { // m found
        if (m % 2 == 0) { // ensure that we find an odd number for k
            k = m + 1;
        } else {
            k = m + 2;
        }
        while (k < 2 * m && k <= s / (2 * m)) {
            if (s / (2 * m) % k == 0 && gcd(k, m) == 1) {
                d = s / 2 / (k * m);
                n = k - m;
                a = d*(m * m - n * n);
                b = 2 * d * n * m;
                c = d * (m * m + n * n);
                found = true;
                break;
            }
            k += 2;
        }
    }
    if (found) {
        break;
    }
}

 

The result looks like

The pythagorean triple is 375, 200, 425, and the sum is 1000
The product is 31875000
Solution took 0 ms

Wrapping up

I am aware that this has been a rather long post. But I needed to explain my self, to really understand the underlying theory of the number theoretical approach. I hope you could follow it and learn something useful , otherwise ask questions and let me know if I can improve parts of the explanation.

As usual you can download the source code

 

From: http://www.mathblog.dk/pythagorean-triplets/