## mukushla Solve in Integers$9^x-3^x=y^4+2y^3+y^2+2y$ one year ago one year ago

1. eliassaab

x=y=1

2. mukushla

and x=0 is trivial sir what is your method

3. eliassaab

I do not have one yet. I can show with Mathematica than no 1<x,y < 1000 are solutions. I do not have a general method yet. I am busy travelling these days.

4. mukushla

np sir :) will work on this later

5. mukushla

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}<z<(2y^{2}+2y+2)^{2}$so only possibility is$z=(2y^2+2y+1)^2=4y^4+8y^3+4y^2+8y+1$it gives$y=1$

6. eliassaab

$9^x- 3^x = y^4 + 2y^3 + y^2 +1\\ 3^x(3^x-1)=y(y+2)(y^2+1)\\$ 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