Solution of problem T6/363 in M&Y


Find all (x,y)\in\mathbb{Z}^2 such that x^{2007}=y^{2007}-y^{1338}-y^{669}+2.

My solution.

Put y^{669}=z then equation has form of x^{2007}=z^3-z^2-z+2(*). Assume that (x,z) is an integer root of (*).

If z=0 then x^{2007}=2, contradiction!

If z=-1 then x^{2007}=1 , so x=1 and y=-1.

If z\in\mathbb{Z}-\{-1,0\} then (z+1)^3-(z^3-z^2-z+2)=4z^2+4z-1=(2z+1)^2-2>0 and (z^3-z^2-z+2)-(z-1)^3=2z^2-4z+3=2(z-1)^2+1>0, therefore (z-1)^3<(x^{669})^3<(z+1)^3, or x^{669}=z. From (*) we have -z^2-z+2=0, or z\in\{1,-2\} , therefore x=1,y=1.

Final, set roots of that equation is \{(1,-1),(1,1)\}.

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