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,hyp otenuse. 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,hyp otenuse. 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.