Problem 39 : Integer right triangles

If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.

{20,48,52}, {24,45,51}, {30,40,50}

For which value of p ≤ 1000, is the number of solutions maximised?

#include <iostream>
 

#include <ctime>

using namespace std;

class PE0039
{
private:
    bool checkRightAngleTriangle(int a, int b, int c);

public:
    int getMaxNumberOfSolutions(int &max_p);
};

bool PE0039::checkRightAngleTriangle(int a, int b, int c)
{
    if (a*a + b*b == c*c)  // a<=c, b<=c
    {
        return
 
 true;
    }
    return false;
}

int PE0039::getMaxNumberOfSolutions(int &max_p)
{
    int max_number = 0;

    for (int p=120; p<=1000; p++)
    {
        int number = 0;
        for(int a=1; a<=p/3; a++)    // a+b+c=p => p>=3a
        {
            for(int b=a; b<=(p-a)/2; b++) // a<=b && b<=p-a-b
 

            {
                if (true == checkRightAngleTriangle(a, b, p-a-b))
                {
#ifdef UNIT_TEST
                    cout << "{" << a << ", " << b << ", " << p-a-b << " )" <<  endl;
#endif
                    number++;
                }
            }
        }
        if (number > max_number)
        {
            max_number = number;
            max_p      = p;
        }
    }

    return max_number;
}

int main()
{
    clock_t start = clock();

    PE0039 pe0039;
    int max_p = 1;
    int max_number = pe0039.getMaxNumberOfSolutions(max_p);

    cout << max_p << " is the number of solutions maximised (";
    cout << max_number << ")." << endl;

    clock_t finish = clock();
    double duration = (double)(finish - start) / CLOCKS_PER_SEC;
    cout << "C/C++ application running time: " << duration << " seconds" << endl;

    return 0;
}