This is C++ programming, please provide detailed programming answers and check whether the output is correct.

Dickson's method: To find positive integer solutions to a2+b2=c2, find positive integers r, s, and t such that r2=2st. Then, (a=r+s,b=r+t,c=r+s+t) form Pythagorean triples. In fact, all Pythagorean triples can be found by this method (by setting r to every possible positive number). Here is an intuitive example: Choose r=6. Then r2/2=18. The three factor-pairs of 18 are: (1,18),(2,9), and (3,6). All three factor-pairs will produce triples using the above equations. s=1,t=18 gives the triple (7,24,25) because a=6+1,b=6+18,c=6+1+18. s=2,t=9 gives the triple (8,15,17) because a=6+2,b=6+9,c=6+2+9. s=3,t=6 gives the triple (9,12,15) because a=6+3,b=6+6,c=6+3+6. Expected Output: The above two examples cannot be efficiently derived using the brutal-force method introduced in Q1. Please use Dickson's method to rewrite. Hint 1: We can iterate the positive integer rfrom the smallest possible value to the upper bound in the outer loop first, as a,b, and c are always larger than r. Note that the step size for iteration could be set to, for example, 2 (the intention is to make sure r is always even and hence r2/2 is an integer). Then compute r2/2 and find all possible factors of this value with an inner loop (think about how to minimize the number of iterations). Lastly, use ifcondition to bound (a=r+s,b=r+t,c=r+s+t) such that a lower bound, b lower bound and c upper bound. Hint 2: First try triples of small ranges as in a) to see if the results stay the same. For large triple based right-angled triangles, the maximal length in test cases will not exceed 50000 , as 500002 reaches the limit of what int can represent (2311)