Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

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

eliassaab Group TitleBest 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.
 2 years ago

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

mukushla Group TitleBest 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\[y=1\]
 2 years ago

eliassaab Group TitleBest 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
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.