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Powers of x and y (Posted on 2017-01-21) Difficulty: 3 of 5
Find all possible couples of distinct integers (x,y) satisfying the equation: x^4 + y = x^3 + y^2

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

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Solution Solution | Comment 1 of 3
Rearrange the equation to become
4x^4 - 4x^3 + 1 = 4y^2 - 4y + 1

The right hand side is the polynomial square (2y-1)^2, so the left side must resolve to the square of an integer for integer (x,y) solutions.

(2x^2 - x)^2 > 4x^4 - 4x^3 + 1 is true for most values of x; x in the interval [-1,1] fail the inequality.

(2x^2 - x - 1)^2 < 4x^4 - 4x^3 + 1 is true for most values of x; x in the interval [0,2/3] fail the inequality.

Then the compound inequality (2x^2 - x - 1)^2 < 4x^4 - 4x^3 + 1 < (2x^2 - x)^2 is true for all x outside the interval [-1,1].

(2x^2 - x - 1)^2 and (2x^2 - x)^2 are consecutive perfect squares.  Then 4x^4 - 4x^3 + 1 is not a square for any x outside the interval [-1,1].  However, 4x^4 - 4x^3 + 1 must be a square for an integer solution.  This implies that all possible integer x are {-1, 0, 1}

If x=-1 then 1+y=-1+y^2, which implies y=2 or y=-1.
If x=0 then 0+y=0+y^2, which implies y=0 or y=1.
If x=1 then 1+y=1+y^2, which implies y=0 or y=1.

Then all the integer solutions are {(-1,-1), (-1,2), (0,0), (0,1), (1,0), (1,1)}.

Edited on January 21, 2017, 3:00 pm
  Posted by Brian Smith on 2017-01-21 14:58:35

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