A community for students.
Here's the question you clicked on:
 0 viewing
Mr.Math
 3 years ago
This group hasn't been active lately. So I'm going to post a set of fun problems bring it back to life.
\[\text{ Fun With Mr.Math #1}\]
Find the integer solutions to the equation
\[9^x3^x=y^4+2y^3+y^2+2y.\]
Mr.Math
 3 years ago
This group hasn't been active lately. So I'm going to post a set of fun problems bring it back to life. \[\text{ Fun With Mr.Math #1}\] Find the integer solutions to the equation \[9^x3^x=y^4+2y^3+y^2+2y.\]

This Question is Closed

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.0why wolframalpha doesn't give solution as x=0 and y=0? http://www.wolframalpha.com/input/?i=9%5Ex3%5Ex%3Dy%5E4%2B2y%5E3%2By%5E2%2B2y+solve+integer

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.3I don't know. I wish it doesn't give any solution at all :D

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0if x=0, y=an imaginary number

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.09^03^0=11=0 and when y=0 then 0+0+0+0=0 same answer, no?

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0factor out y, you'll end up with \(\ (y^2+2y)(y^2+1)=9^x3^x \) when x=0, \(\ y^2=1, y=\sqrt{1} \)

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0but \(\ y^2+2y=0 => y=2 \)

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.0y(y+2)(y^2+1)=0 so y=0 and y=2 no?

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0please write full statements lol, you're getting me lost by ... no? :P

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.0you write \[(y^2+2y)(y^2+1)=9^x3^x\] if x=0 \[(y^2+2y)(y^2+1)=0\] \[y(y+2)(y^2+1)=0\] so \[y=0\text{ and }y+2=0\text{ and }y^2+1=0\] so y=0 y=2 and no real solutions

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0=> my solutions is wrong? not sure... i guess i'll wait for to see the correct one :D

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.0i don't care about solution because i dont have a clue how to solve but x=0 and y=0 is clearly true and u didn't prove me that it's wrong lol

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0:P y=2 and y=0 both work for x=0 and are true, but i wonder if they're the right solutions lol, Mr. Math, what are the right solutions?

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.0so why wolframalpha does not show y=0?

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0it isn't as smart as u ;)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3Here are my thoughts so far: First lets simplify the left hand side:\[9^x3^x=(3^2)^x3^x=3^{2x}3^x=3^x(3^x1)\]Then lets simplify the right hand side:\[y^4+2y^3+y^2+2y=y(y+2)(y^2+1)\] We therefore have:\[3^x(3^x1)=y(y+2)(y^2+1)\]Now the trivial solution found above is x=0 which leads to y=0 or y=2 (since we are only interested in integer solutions). Another solution could also be \(x=\infty\) and y=0 or 2 but I assume we can rule out infinities. Now lets rearrange the equation to:\[y^2+1=\frac{3^x(3^x1)}{y(y+2)}\]This means \(y(y+2)\) must evenly divide into \(3^x(3^x1)\). Notice that \(3^x\) is always an odd number and \(3^x1\) is always an even number. Case 1: If y is an even number, then y+2 will also be even. Therefore, if y is even, then \(y(y+2)\) must evenly divide into \(3^x1\). So let \(y=2m\) to get:\[\frac{3^x1}{4m(m+1)}\hspace{2cm}\text{must be an integer}\] Case 2: If y is an odd number, then y+2 will also be odd. Therefore, if y is odd, then \(y(y+2)\) must evenly divide into \(3^x\). So let \(y=2m+1\) to get:\[\frac{3^x}{(2m+1)(2m+3)}\hspace{2cm}\text{must be an integer}\]

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3Since 3 is a prime number, case 2 can be rejected because \((2m+1)(2m+3)\) can only evenly divide into \(3^x\) if \((2m+1)(2m+3)=3^k\) which is impossible. So we are left with just Case 1 to solve.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3In case 1, if m>1, then \(4m(m+1)\) will always contain a product of an even number and an odd number. However, we know the numerator \(3^x1\) is always even, therefore we can reject all values of m>1. So case 1 reduces to m=1 which gives y=2 (since \(y=2m\) for this case). Substituting this back into the original equation gives:\[\begin{align} 3^x(3^x1)&=y(y+2)(y^2+1)\\ &=2(2+2)(2^2+1)\\ &=2*4*5\\ &=40 \end{align}\]which does not give an integer solution for x. Therefore (I believe) the only solutions are: 1) x=0 and y=0 or 2 2) x=\(\infty\) and y=0 or 2

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.3Great job everyone, some solution(s) is/are still missing though.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3hmmm  I rechecked my work and found some flaws in my logic. I have now proved that for y>0 y must be odd. And, by trialanderror I have found one more solution: 3) x=1 and y=1 but I have not been able to prove that this is the only other solution :(

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0its immediate that \(x\ge0\) letting \(x=0\) gives 2 solution \[(x,y)=(0,0)\ , \ (0,2)\]suppose that \(x>0\) using asnaseer factoring\[3^x(3^x1)=y(y+2)(1+y^2)\]RHS is divisible by \(3^x\)\[y \equiv 0,1,2 \ \ \text{mod} \ 3 \\ y^2 \equiv 0,1 \ \ \text{mod} \ 3 \\ 1+y^2 \equiv 1,2 \ \ \text{mod} \ 3 \]so \(y(y+2)\) is divisible by \(3^x\) but we know that \(\gcd(y,y+2)=1,2\) so exactly one of \(y\) or \(y+2\) is divisible by \(3^x\). case 1 : \(y=3^xm\) and \(m\neq0\) and \(m\) is not a multiple of 3\[3^x1=m(3^xm+2)(1+m^23^{2x})\]RHS is greater than LHS in magnitude so no solution from here case 2 : \(y+2=3^xm\) and \(m\neq0\) and \(m\) is not a multiple of 3\[3^x1=(3^xm2)m(5+m^23^{2x}4m3^x)\]\[3^x1=m(3^xm2)(5+(3^xm^24m)3^x)\]again RHS is greater than LHS in magnitude unless \(x=1\) so one solution from here\[(x,y)=(1,1)\]

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0actually for case 2 if \(x>1\)\[m(3^xm2)(5+(3^xm^24m)3^x)>(5+(3^xm^24m)3^x)>5+3^x\]

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0only solutions\[(x,y)=(0,0)\ ,\ (0,2)\ , \ (1,1)\]
Ask your own question
Sign UpFind more explanations on OpenStudy
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.