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mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2and x=0 is trivial sir what is your method

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1I 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.

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2np sir :) will work on this later

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2clearly \(x\ge0 \) for x=0 we have 2 solution (x,y)=(0,0),(0,2) x>0\[3^{2x}3^x=y^4+2y^3+y^2+2y\]\[4(3^{2x}3^x)+1=(2\times3^x1)^2=4(y^4+2y^3+y^2+2y)+1\]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\[y=1\]

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.1\[ 9^x 3^x = y^4 + 2y^3 + y^2 +1\\ 3^x(3^x1)=y(y+2)(y^2+1)\\ \] 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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