It is obviously true that $x>y$ . You can say $x=y+d$ and rewrite the expression. After that find roots and it should finish.

Omeredip wrote: It is obviously true that x>y . You can say x=y+d and rewrite the expression. After that find roots and it should finish. I don't think that would work as you would still be setting 2 variables, and I am not sure if it would turn out nice. Maybe you could specifically show your solution so I understand...anyways I think due to the symmetry in this problem, there must be a much better way... Edit: I found this on stack exchange, the solution is pretty much using the same idea as above, but I feel utilizes the symmetry better...then again it isn't too nice either but....

https://artofproblemsolving.com/community/c6h348875

vatatmaja wrote: Omeredip wrote: It is obviously true that x>y . You can say x=y+d and rewrite the expression. After that find roots and it should finish. I don't think that would work as you would still be setting 2 variables, and I am not sure if it would turn out nice. Maybe you could specifically show your solution so I understand...anyways I think due to the symmetry in this problem, there must be a much better way... Edit: I found this on stack exchange, the solution is pretty much using the same idea as above, but I feel utilizes the symmetry better...then again it isn't too nice either but.... In my last message, stage of finding roots was wrong .Let me explain. We can say $x>y$ and these numbers are natural numbers so let we use $y+d=x$ (d is integer) hence $(y+d)^3-y^3=y^2+yd+61$ $(3d-1).y^2+d(3d-1)y+d^3=61$ We found that $d \leq 3$ If $d=1$ , $2y^2+2y+1=61$ 》$y=5$ and $x=6$ If $d=2$ , $5y^2+10y+8=61$ has no solution If $d=3$ , $8y^2+24y+27=61$ has no solution So $(x,y)=(6,5)$ This problem is in http://www.dr.com.tr/Kitap/Analiz-ve-Cebirde-Ilginc-Olimpiyat-Problemleri-ve-Cozumleri/Ilham-Aliyev/Bilim/Matematik/urunno=0000000546243