Find Pythagorean triplet for which a + b + c = 1000

Question

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a² + b² = c²

For example, 3² + 4

Answer 1

I'm afraid ^ doesn't do what you think it does in C. Your best bet is to use a*a for integer squares.

Answer 2

Here's a solution using Euclid's formula (link).

Let's do some math: In general, every solution will have the form

a=k(x²-y²)
b=2kxy
c=k(x²+y²)

where k, x and y are positive integers, y < x and gcd(x,y)=1 (We will ignore this condition, which will lead to additional solutions. Those can be discarded afterwards)

Now, a+b+c= kx²-ky²+2kxy+kx²+ky²=2kx²+2kxy = 2kx(x+y) = 1000

Divide by 2: kx(x+y) = 500

Now we set s=x+y: kxs = 500

Now we are looking for solutions of kxs=500, where k, x and s are integers and x < s < 2x. Since all of them divide 500, they can only take the values 1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, 500. Some pseudocode to do this for arbitrary n (it and can be done by hand easily for n=1000)

If n is odd
  return "no solution"
else
  L = List of divisors of n/2
for x in L
  for s in L
    if x< s <2*x and n/2 is divisible by x*s
      y=s-x
      k=((n/2)/x)/s      
      add (k*(x*x-y*y),2*k*x*y,k*(x*x+y*y)) to list of solutions
sort the triples in the list of solutions
delete solutions appearing twice
return list of solutions

You can still improve this:

• x will never be bigger than the root of n/2
• the loop for s can start at x and stop after 2x has been passed (if the list is ordered)

For n = 1000, the program has to check six values for x and depending on the details of implementation up to one value for y. This will terminate before you release the button.

Answer 3

There is a quite dirty but quick solution to this problem. Given the two equations

a*a + b*b = c*c

a+b+c = 1000.

You can deduce the following relation

a = (1000*1000-2000*b)/(2000-2b)

or after two simple math transformations, you get:

a = 1000*(500-b) / (1000 - b)

since a must be an natural number. Hence you can:

for b in range(1, 500):
    if 1000*(500-b) % (1000-b) == 0:
        print b, 1000*(500-b) / (1000-b) 

Got result 200 and 375.

Good luck

Answer 4

I know this question is quite old, and everyone has been posting solutions with 3 for loops, which is not needed. I got this solved in O(n), by **equating the formulas**; **a+b+c=1000 and a^2 + b^2 = c^2**

So, solving further we get;

a+b = 1000-c

(a+b)^2 = (1000-c)^2

If we solve further we deduce it to;

a=((50000-(1000*b))/(1000-b)). We loop for "b", and find "a".

Once we have "a" and "b", we get "c".

public long pythagorasTriplet(){
    long a = 0, b=0 , c=0;

    for(long divisor=1; divisor<1000; divisor++){
        if( ((500000-(1000*divisor))%(1000-divisor)) ==0){
            a = (500000 - (1000*divisor))/(1000-divisor);
            b = divisor;
            c = (long)Math.sqrt(a*a + b*b);
            System.out.println("a is " + a + " b is: " + b + " c is : " + c);
            break;
        }
    }
    return a*b*c;
}

Answer 5

Euclid method gives the perimeter to be m(m+n)= p/2 where m> n and the sides are m^2+n^2 is the hypotenuse and the legs are 2mn and m^2-n^2.thus m(m+n)=500 quickly gives m= 20 and n=5. The sides are 200, 375 and 425. Use Euclid to solve all pythorean primitive questions.

Answer 6

As there are two equations (a+b+c = 1000 && aˆ2 + bˆ2 = cˆ2) with three variables, we can solve it in linear time by just looping through all possible values of one variable, and then we can solve the other 2 variables in constant time.

From the first formula, we get b=1000-a-c, and if we replace b in 2nd formula with this, we get c^2 = aˆ2 + (1000-a-c)ˆ2, which simplifies to c=(aˆ2 + 500000 - 1000a)/(1000-a).

Then we loop through all possible values of a, solve c and b with the above formulas, and if the conditions are satisfied we have found our triplet.

    int n = 1000;

    for (int a = 1; a < n; a++) {
        int c = (a*a + 500000 - 1000*a) / (1000 - a);
        int b = (1000 - a - c);

        if (b > a && c > b && (a * a + b * b) == c * c) {
            return a * b * c;
        }
    }