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mukushlaBest ResponseYou've already chosen the best response.2
and x=0 is trivial sir what is your method
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
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.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
np sir :) will work on this later
 one year ago

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

eliassaabBest 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
 one year ago
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