Mr.Math
  • Mr.Math
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 \[9^x-3^x=y^4+2y^3+y^2+2y.\]
Meta-math
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
why wolframalpha doesn't give solution as x=0 and y=0? http://www.wolframalpha.com/input/?i=9%5Ex-3%5Ex%3Dy%5E4%2B2y%5E3%2By%5E2%2B2y+solve+integer
Mr.Math
  • Mr.Math
I don't know. I wish it doesn't give any solution at all :-D
sasogeek
  • sasogeek
if x=0, y=an imaginary number

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
why?
anonymous
  • anonymous
9^0-3^0=1-1=0 and when y=0 then 0+0+0+0=0 same answer, no?
sasogeek
  • sasogeek
factor out y, you'll end up with \(\ (y^2+2y)(y^2+1)=9^x-3^x \) when x=0, \(\ y^2=-1, y=\sqrt{-1} \)
sasogeek
  • sasogeek
but \(\ y^2+2y=0 => y=-2 \)
anonymous
  • anonymous
y(y+2)(y^2+1)=0 so y=0 and y=-2 no?
sasogeek
  • sasogeek
please write full statements lol, you're getting me lost by ... no? :P
anonymous
  • anonymous
you write \[(y^2+2y)(y^2+1)=9^x-3^x\] 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
  • sasogeek
=> my solutions is wrong? not sure... i guess i'll wait for to see the correct one :D
anonymous
  • anonymous
i 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
  • sasogeek
: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?
anonymous
  • anonymous
so why wolframalpha does not show y=0?
sasogeek
  • sasogeek
it isn't as smart as u ;)
asnaseer
  • asnaseer
Here are my thoughts so far: First lets simplify the left hand side:\[9^x-3^x=(3^2)^x-3^x=3^{2x}-3^x=3^x(3^x-1)\]Then lets simplify the right hand side:\[y^4+2y^3+y^2+2y=y(y+2)(y^2+1)\] We therefore have:\[3^x(3^x-1)=y(y+2)(y^2+1)\]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:\[y^2+1=\frac{3^x(3^x-1)}{y(y+2)}\]This means \(y(y+2)\) must evenly divide into \(3^x(3^x-1)\). Notice that \(3^x\) is always an odd number and \(3^x-1\) 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 \(3^x-1\). So let \(y=2m\) to get:\[\frac{3^x-1}{4m(m+1)}\hspace{2cm}\text{must be an integer}\] 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
  • asnaseer
Since 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
  • asnaseer
In 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 \(3^x-1\) 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:\[\begin{align} 3^x(3^x-1)&=y(y+2)(y^2+1)\\ &=2(2+2)(2^2+1)\\ &=2*4*5\\ &=40 \end{align}\]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
  • Mr.Math
Great job everyone, some solution(s) is/are still missing though.
asnaseer
  • asnaseer
hmmm - I re-checked my work and found some flaws in my logic. I have now proved that for y>0 y must be odd. And, by trial-and-error 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 :(
anonymous
  • anonymous
its immediate that \(x\ge0\) letting \(x=0\) gives 2 solution \[(x,y)=(0,0)\ , \ (0,-2)\]suppose that \(x>0\) using asnaseer factoring\[3^x(3^x-1)=y(y+2)(1+y^2)\]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\[3^x-1=m(3^xm+2)(1+m^23^{2x})\]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\[3^x-1=(3^xm-2)m(5+m^23^{2x}-4m3^x)\]\[3^x-1=m(3^xm-2)(5+(3^xm^2-4m)3^x)\]again RHS is greater than LHS in magnitude unless \(x=1\) so one solution from here\[(x,y)=(1,1)\]
anonymous
  • anonymous
actually for case 2 if \(x>1\)\[|m(3^xm-2)(5+(3^xm^2-4m)3^x)|>|(5+(3^xm^2-4m)3^x)|>5+3^x\]
anonymous
  • anonymous
only solutions\[(x,y)=(0,0)\ ,\ (0,-2)\ , \ (1,1)\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.