Solve in Integers\[9^x-3^x=y^4+2y^3+y^2+2y\]

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Solve in Integers\[9^x-3^x=y^4+2y^3+y^2+2y\]

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x=y=1
and x=0 is trivial sir what is your method
I do not have one yet. I can show with Mathematica than no 1

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np sir :) will work on this later
clearly \(x\ge0 \) for x=0 we have 2 solution (x,y)=(0,0),(0,-2) x>0\[3^{2x}-3^x=y^4+2y^3+y^2+2y\]\[4(3^{2x}-3^x)+1=(2\times3^x-1)^2=4(y^4+2y^3+y^2+2y)+1\]so\[z=4(y^4+2y^3+y^2+2y)+1=4y^4+8y^3+4y^2+8y+1\]must be a complete square but\[(2y^{2}+2y)^{2}
\[ 9^x- 3^x = y^4 + 2y^3 + y^2 +1\\ 3^x(3^x-1)=y(y+2)(y^2+1)\\ \] One can see that the only integer solutions (x,y) with \(x \le 1\) are (0,-2),(0,0) and (1,1). We claim that these are the only one. Suppose the contrary that (x,y) is an integer solution with \(x\ge 2\) and \(y\notin \{-2,0,1\}\). It is easy to see that \(y^2+1 \) is not a multiple of 3. Since gcd(y,y+2) is either 1 or 2 it follows that \( 3^x\) divides either y or y+2 If \( 3^x\) divides y and since \(y\ne 0\) we have \(y\ge 3^x \) hence \[ y(y+2)(y^2+1)\ge y^3 \ge 27^x > 9^x -3^x \] Contradiction. If If \( 3^x\) divides y+2 Write \(y= 3^x k -2 \) with \(k\ne 0\). Hence \( 5k^2 \ge 4k +1 \). As \(x\ge 2\) \[ 9^x k^2 = 3^x 3^x k^2> 3^x 5k^2 \ge 3^x(4k+1) \] Therefore \[ y^2+1 = 9^x k^2 -3^x(4k) +5 > 3^x \] Finally since\( |y|\ge 1 \) and \( y +2 \ge 3^x\), we have \[ y(y+2)(y^2+1)> 1 \times 3^x\times 3^x=9^x> 9^x -3^x \] Contradiction

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