squared numbers is in range

[Karalis]

Hey,
Won't lie this is a teachers assigment but i became kinda sad, cause i have no idea how to do last part of it.
I'm supposed to find all pairs of positive integers whose squares add up to a number in the given range.

Well im stuck with that...
I need to sum all the number squares of for example:
1 + 1 = 2
1 + 2 = 5
1 + 3 = 10
1 + 4 =17
2 + 2 = 8
and so on...
And when user enter a range of 1-10 for example...
I need to print out the numbers that makes sum that is in the users range if that makes sence.

range 1-10
1 + 1 = 2
2+2 =8
1 + 3 = 10
1+2 = 5

range 5-15
1+9 = 10
1+2 = 5
2+3 = 13

I try do this with while loop, had 0 success. I did this on paper but i have no idea how to start to write it.

12 + 12 = 2     print to screen and increment right square
12 + 22 = 5     print to screen and increment right square
12 + 32 = 10    print to screen and increment right square
12 + 42 = 17    17 is > 10, so increment left square and right square = 1
22 + 12 = 5     print to screen and increment right square
22 + 22 = 8     print to screen and increment right square
22 + 32 = 13    13 is > 10, so increment left square and right square = 1
32 + 12 = 10    print to screen and increment right square
32 + 22 = 13    13 is > 10, so increment left square and right square = 1
42 + 12 = 17    The program would stop here

  int main(){
     	int * intervals, * kvadrats;
		int maslielums,m,n,i = 0;
		int mazakais = 0;
     	bool parbaude1;
     	char at = 'j';
    
     	while(at == 'j'){
    //checks if m is entered correct
      do {
           parbaude1 = false;
           cout<<"Intervala m veribu: ";
           cin >> m;
           if (!cin.good()) {
           cout << "Ievadiet tikai naturalu skaitli!" << endl;
           cin.clear();
           cin.ignore(256,'\n');
           parbaude1 = true;
         }//cannot be negative
             if(m<0) {
             cout << "Skaitlis nedrikst but negativs: " << endl;
             parbaude1 = true;
           }
    } while(parbaude1 == true);
    
    // checks if n is entered correctly
    do {
           parbaude1 = false;
           cout<<"Intervala n veribu: ";
           cin >> n;
           if (!cin.good()) {
           cout << "Ievadiet tikai naturalu skaitli!" << endl;
           cin.clear();
           cin.ignore(256,'\n');
           parbaude1 = true;
         }// nedrikst but negativs!
             if(n<0) {
             cout << "Skaitlis nedrikst but negativs: " << endl;
             parbaude1 = true;
           }
    } while(parbaude1 == true);
     
     // Check which of the number n or m is smaller.
	  if (m>n){
      intervals = new int[m-n+1];
      maslielums = m-n; // gives out users range
      mazakais = n; // smallest number that user entered
      }
          else{ 
      intervals = new int[n-m+1];
      maslielums = n-m; // gives out users range
      mazakais = m; // smallest number that user entered
        }
    //fils out range with numbers  
    for (int i=0; i<=maslielums; i++){
    intervals[i] = mazakais;
    mazakais++; 
}
    
    
		
   cout << endl << "Ja velies atkartot programmu ievadi 'j',bet ja beigt tad ievadiet jebko citu!"<< endl; 
   cin >> at;
   delete[] intervals;
}
return 0;
}

[Karalis]

1² + 1² = 2 print to screen and increment right square
1² + 2² = 5 print to screen and increment right square
1² + 3² = 10 print to screen and increment right square
1² + 4² = 17 17 is > 10, so increment left square and right square = 1
2² + 1²= 5 print to screen and increment right square
2² + 2² = 8 print to screen and increment right square
2² + 3² = 13 13 is > 10, so increment left square and right square = 1
3² + 1² = 10 print to screen and increment right square
3²+ 2² = 13 13 is > 10, so increment left square and right square = 1
4²+ 1²= 17 The program would stop here
If the user enters 1-15 for example.

[2kaud]

i have no idea how to start to write it.

The basic algorithm is very simple. Consider

	for (int x = start; x <= fin; ++x)
		for (int y = start; y <= fin; ++y) {
			int z = x * x + y * y;
			if (z >= n && z <= m)
				cout << x << " " << y << " " << z << endl;
		}

where n is the lower range limit and m is the upper range limit. start and fin are the start and finish numbers for the loops - which need to be determined. Also this algorithm can be improved for efficiency once z is greater than m. But I'll leave this an an exercise!