Find Pythagorean triplet for which a + b + c = 100

Question

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.

Source: http://projecteuler.net/index.php?section=problems&id=9

I tried but didn't know where my code went wrong. Here's my code in C:

#include <math.h>
#include <stdio.h>
#include <conio.h>


void main()
{
    int a=0, b=0, c=0;
    int i;
    for (a = 0; a<=1000; a++)
    {
        for (b = 0; b<=1000; b++)
        {
            for (c = 0; c<=1000; c++)
            {
                if ((a^(2) + b^(2) == c^(2)) && ((a+b+c) ==1000)))
                    printf("a=%d, b=%d, c=%d",a,b,c);
            }
        }
    }
getch();    
}

Answer 1

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 2

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;
        }
    }

Answer 3

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 4

#include <stdio.h>

int main() // main always returns int!
{
 int a, b, c;
 for (a = 0; a<=1000; a++)
 {
  for (b = a + 1; b<=1000; b++) // no point starting from 0, otherwise you'll just try the same solution more than once. The condition says a < b < c.
  {
   for (c = b + 1; c<=1000; c++) // same, this ensures a < b < c.
   {
    if (((a*a + b*b == c*c) && ((a+b+c) ==1000))) // ^ is the bitwise xor operator, use multiplication for squaring
     printf("a=%d, b=%d, c=%d",a,b,c);
   }
  }
 }
 return 0;
}

Haven't tested this, but it should set you on the right track.