Using Pythagoras, we get
(5a+12b)2+(12a+5b)2 = (13a+kb)2
Solving for k:
The second solution is negative and does not fit the requirements that k is a positive integer.
We are required to find pairs of positive integers a and b such that k is also a positive integer.
I have not been able to find an analytical solution of the integer problem, although the minimum value of k appears to be ten, with the minimum values of a and b being 69 and 20.
Thus the triplet (a,b,k) of (69,20,10) is the minimal solution.
The solution for k=11 is (24,23,11), and for k=12, (25,72,12). Other solutions exist for the same k where a and b are multiples of the basic case.
the answer is 3==D