Let #5a+12b# and #12a+5b# be the side lengths of a right-angled triangle and #13a+kb# be the hypotenuse, where #a#, #b# and #k# are positive integers. How do you find the smallest possible value of #k# and the smallest values of #a# and #b# for that #k#?

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Answer 1 · 2018-02-25T01:45:57

#k = 10#, #a=69#, #b=20#

By Pythagoras' theorem, we have:

#(13a+kb)^2 = (5a+12b)^2+(12a+5b)^2#

That is:

#169a^2+26kab+k^2b^2 = 25a^2+120ab+144b^2+144a^2+120ab+25b^2#

#color(white)(169a^2+26kab+k^2b^2) = 169a^2+240ab+169b^2#

Subtract the left hand side from both ends to find:

#0 = (240-26k)ab + (169-k^2)b^2#

#color(white)(0) = b((240-26k)a+(169-k^2)b)#

Since #b > 0# we require:

#(240-26k)a+(169-k^2)b = 0#

Then since #a, b > 0# we require #(240-26k)# and #(169-k^2)# to have opposite signs.

When #k in [1, 9]# both #240-26k# and #169-k^2# are positive.

When #k in [10, 12]# we find #240-26k < 0# and #169-k^2 > 0# as required.

So the minimum possible value of #k# is #10#.

Then:

#-20a+69b = 0#

Then since #20# and #69# have no common factor larger than #1#, the minimum values of #a# and #b# are #69# and #20# respectively.