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 \[9^x-3^x=y^4+2y^3+y^2+2y.\]

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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 \[9^x-3^x=y^4+2y^3+y^2+2y.\]

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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
I don't know. I wish it doesn't give any solution at all :-D
if x=0, y=an imaginary number

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why?
9^0-3^0=1-1=0 and when y=0 then 0+0+0+0=0 same answer, no?
factor out y, you'll end up with \(\ (y^2+2y)(y^2+1)=9^x-3^x \) when x=0, \(\ y^2=-1, y=\sqrt{-1} \)
but \(\ y^2+2y=0 => y=-2 \)
y(y+2)(y^2+1)=0 so y=0 and y=-2 no?
please write full statements lol, you're getting me lost by ... no? :P
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
=> my solutions is wrong? not sure... i guess i'll wait for to see the correct one :D
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
: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?
so why wolframalpha does not show y=0?
it isn't as smart as u ;)
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:\[\frac{3^x}{(2m+1)(2m+3)}\hspace{2cm}\text{must be an integer}\]
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.
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
Great job everyone, some solution(s) is/are still missing though.
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 :(
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\[(x,y)=(1,1)\]
actually for case 2 if \(x>1\)\[|m(3^xm-2)(5+(3^xm^2-4m)3^x)|>|(5+(3^xm^2-4m)3^x)|>5+3^x\]
only solutions\[(x,y)=(0,0)\ ,\ (0,-2)\ , \ (1,1)\]

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