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December 18, 2014

December 18, 2014

Posted by **m** on Sunday, July 19, 2009 at 2:20am.

- math -
**MathMate**, Sunday, July 19, 2009 at 11:15amUsing 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.

- math -
**ping yau peng shu (not)**, Sunday, September 13, 2009 at 12:04amthe answer is 3==D

- math -
**werp**, Sunday, September 13, 2009 at 12:05ammaths... keewwllllll

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