let 5a+12b and 12a+5b be the sides length of a right-angled triangle and let 13a+kb be the hypotenuse,l where a,b and k are positive integers. find the smallest possible value of k and the smallest values of a and b for that k.
we know that
(5a+12b)^2 + (12a+5b)^2 = (13a+kb)^2
25a^2+120ab+144b^2 + 144a^2+120ab+25b^2 = 169a^2 + 26kab + k^2b^2
169a^2+240ab+169b^2 = 169a^2+26kab+k^2b^2
(240-26k)ab + (169-k^2)b^2 = 0
b((240-26k)a+(169-k^2)b) = 0
Since a and b are positive, either
249-26k < 0 or 169-k^2 < 0
So, if k=10, we have
-20ab + 69b^2 = 0
b(69b-20a) = 0
Looks like a=69, b=20 will work