Let #a# and #b# be positive integers such that #ab + 1# divides #a^2 + b^2#. Show that #(a^2 + b^2) / (ab + 1)# is the square of an integer. ?

Question

4932 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-03T18:24:41

See below.

Supposing that

#(a^2+b^2)/(a b + 1) = k# then

#a^2-kb a + b^2-k = 0# solving for #a#

#a = (k bpm sqrt(k^2b^2-4(b^2-k)))/2#

all the #b#'s obeying #b^2=k# and the #a#'s obeying

#a=kb# are solutions. So

#((b,a,k),(1,1,1),(2,8,4),(3,27,9),(4,64,16),(5,125,25),(6,216,36),(cdots,cdots,cdots))#

The question follows. Are those all solutions?