Solving right triangle w/Pythagorean theorem

[inferi9]

hi everyone I am new here and I have this C++ program that I have to write but it keep given me nothing useful. here is the question:


A right triangle can have sides that are all integers. A set of three integer values for the sides of a right triangle is called a Pythagorean triple. These three sides must satisfy the following relationship:
(side1)^2 + (side2)^2 = (hypotenuse)^2
Output all Pythagorean triples for side1, side2, and hypotenuse all no longer 100.

For example:
side1 side2 hypotenuse
3 4 5


you have to do it using nested for only and print the result.

//A program that cheakes A right triangle 
#include <iostream>
using namespace std;
int main ()
{
    //Declare variables.
    int side1, side2, hypotenuse,i;
 
 
 
    for (i=1;i<=100;i++)
    {
        for (side1=1;side1<=100;side1++)
 
 
 
    for (side2=1;side2<=100;side2++)
 
 
        for (hypotenuse=1;hypotenuse<=100;hypotenuse++)
 
 
 
 
 
        if ((hypotenuse*hypotenuse)==((side1*side1)+(side2*side2)))
 
 
            cout<<side1<<side2<<hypotenuse<<endl;
 
 
    }
 
 
 
 
            return 0;
 
 
    }

[weaknessforcats]

What is your question?

Have you tried this code?

[inferi9]

my question is that I have to write a loop for side1 and a loop foe side two and a loop for the hypotenuse. and the three must be for loop. the must stopes when this equation: (side1)^2 + (side2)^2 = (hypotenuse)^2
is found and the print all side1,side2 and hypotenuse. i tried this code but it gives me an infinite loop.
can anyone help?
thank you

[Ganon11]

Nothing in that code should give you an infinite loop - it simply takes a very long time to execute. One thing you can do to shorten this time is to remove the outermost loop (the one with i controlling it), as this contributes nothing to your code, but multiplies the executions done by 100. Still, your code is doing 100 * 100 * 100 = 1,000,000 executions, so give it some time before deciding it's an infinite loop.

[inferi9]

i waited and it stopped but it gave me a very crazy numbers i want only to execute the side1,side2,hypotenuse. that satisfay the equation (side1)^2 + (side2)^2 = (hypotenuse)^2

[inferi9]

i used the if statment inside and outside the loop and it is no good

[JosAH]

Have a look at this page and pay special attention to formula #11.

kind regards,

Jos

[Ganon11]

i waited and it stopped but it gave me a very crazy numbers i want only to execute the side1,side2,hypotenuse. that satisfay the equation (side1)^2 + (side2)^2 = (hypotenuse)^2

I took your code and did 2 things to it:

1) I removed the first for...loop, as it was unnecessary.
2) You typed in side2 with some spaces in the middle, and I removed those spaces so it actually compiled.

After this, your program worked like a charm, giving me all the pythagorean triples I needed.

[inferi9]

I took your code and did 2 things to it:

1) I removed the first for...loop, as it was unnecessary.
2) You typed in side2 with some spaces in the middle, and I removed those spaces so it actually compiled.

After this, your program worked like a charm, giving me all the pythagorean triples I needed.

I removed the first for loop but I do not really get that there are spaces in side2.
the other thing is do I have to put one of the loops as the first loop and put the others in a btraket { }.

[Ganon11]

Show us the code you are currently working with.

[inferi9]

here it is what I am working on:

//A program that cheakes A right triangle 
#include <iostream>
using namespace std;
int main ()
{
    //Declare variables.
    int side1, side2, hypotenuse;
 
    for (side1=1;side1<=100;side1++)
 
 
    for (side2=1;side2<=100;side2++)
 
 
    for (hypotenuse=1;hypotenuse<=100;hypotenuse++)
 
 
 
    if ((hypotenuse*hypotenuse)==((side1*side1)+(side2*side2)))
 
 
            cout<<side1<<side2<<hypotenuse<<endl;
 
 
 return 0;
 
 
    }

and it givese avery crazy numbers.

[Ganon11]

The only thing I can see is that you are printing each number right on top of the other, with no spaces in between. So, for example, the best known pythagorean triple, (3, 4, 5), will print as 345. Put some spaces in there too get separate numbers.