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A cubic equation (Posted on 2023-09-14) Difficulty: 3 of 5
Solve in positive integers:


Source: Bulgarian Math. Competition

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

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Solution Solution | Comment 1 of 2
For most values of y, the following compound inequality is true:
y^3 < y^3+2y^2+1 < (y+1)^3
If y satisfies this inequality then the expression y^3+2y^2+1 cannot be a perfect cube because it is strictly between two consecutive cubes.

The places when this compound inequality fails to hold are when y^3 >= y^3+2y^2+1 OR y^3+2y^2+1 >= (y+1)^3.
If y^3 >= y^3+2y^2+1 then 0 >= 2y^2+1, but 2y^2+1 is always positive, so no possible cases here.
If y^3+2y^2+1 >= (y+1)^3 then 0 >= y^2+3y.  The integers which satisfy this last inequality are 0, -1, -2, and -3

Then for each case:
y=0 -> y^3+2y^2+1=1 -> x=1
y=-1 -> y^3+2y^2+1=2 -> not a perfect cube 
y=-2 -> y^3+2y^2+1=1 -> x=1 
y=-3 -> y^3+2y^2+1=-8 -> x=-2
So there are three integer solutions: (x,y) = (1,0), (1,-2) and (-2, -3).  However none of these are pairs of positive integers.

  Posted by Brian Smith on 2023-09-14 23:59:54
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