Posted by m on Sunday, July 19, 2009 at 2:20am.
Using Pythagoras, we get
(5a+12b)^{2}+(12a+5b)^{2} = (13a+kb)^{2}
Solving for k:
k=(sqrt(169*b^2+240*a*b+169*a^2)-13*a)/b
or
k=-(sqrt(169*b^2+240*a*b+169*a^2)+13*a)/b
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
maths... keewwllllll