Find the positive integer solutions to the equation #x^3-y^3=x y+61#?

Question

3314 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-01T18:41:04

See below.

With #p= 61#

Making the transformation

#{(x+y=a),(x-y=b):}#

equivalent to a rotation of #pi/4# clockwise with the purpose of cross product #xy# elimination, we obtain

#b^2 + b^3 + a^2 (3 b-1) - 4 p=0# then

#a^2=(4p-b^3-b^2)/(3b-1)#

Here #a^2 > 0# and #p=61# then

#b lt 6#

Now evaluating #(4*61-b^3-b^2)/(3b-1)# for #b in {1,2,cdots,5}# we have

#{121, 232/5, 26, 164/11, 47/7}#

The only feasible outcome is #121=11^2# for #b=1#

so we have

#{(x+y=11),(x-y=1):}#

giving

#x=6# and #y = 5#