Solve in the natural numbers the equation

9^{x}-3^{x}=y^{4}+2y^{3}+y^{2}+2y^{}

Start by multiplying both sides by 4 and adding one: 4*9^x - 4*3^x + 1 = 4y^4 + 8y^3 + 4y^2 + 8y + 1

The left hand side is a perfect square: 4*9^x - 4*3^x + 1 = (2*3^x - 1)^2.

The right side can be squeezed between a pair of perfect squares: (2y^2 + 2y - 1)^2 < 4y^4 + 8y^3 + 4y^2 + 8y + 1 < (2y^2 + 2y + 1)^2

For the equality to hold either the expression in the middle of the inequality equals (2y^2+2y)^2 or one part of the compound inequality fails.

In the first case then 4y^4 + 8y^3 + 4y^2 + 8y + 1 = (2y^2+2y)^2. This resolves to y=-1/8, which is not an integer solution.

In the second case either (2y^2 + 2y - 1)^2 > 4y^4 + 8y^3 + 4y^2 + 8y + 1 OR 4y^4 + 8y^3 + 4y^2 + 8y + 1 > (2y^2 + 2y + 1)^2. The first inequality resolves to 0 > 4y^2+12y, which is true on the interval [-3,0]. The second inequality resolves to 0 > 4y^2-4y, which is true on the interval [0,1]. The union of the two intervals is [-3,1], which has exactly one natural number in the interval: y=1.

Then substituting y=1 in the original equation makes 9^x-3^x = 6, for which x=1 is a solution. Then the solutions of the original equation over the natural numbers is the single ordered pair **(x,y) = (1,1)**.