Perfect Square
Find all integers x with the property that $2x^3+9$ is perfect square!
- Fahim Shahriar
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- Joined:Sun Dec 18, 2011 12:53 pm
Re: Perfect Square
Let $2x^3+9=y^2$
$2x^3=(y+3)(y-3)$
Surely $y$ is odd. Then $(y+3)$ and $(y-3)$ both will be even. Take $(y-3)=2m$.
$2x^3=2m(2m+6)$
$x^3=2m(m+3)$
Note that $x$ is even. So both of $m$ and $(m+3)$ are powers of 2.
Suppose, $m=2^a$ and $m+3=2^b$
$2^b-2^a=3$
It implies $a$ is 0. So $m=1$. Only solution for $x$ is $2$.
$2x^3=(y+3)(y-3)$
Surely $y$ is odd. Then $(y+3)$ and $(y-3)$ both will be even. Take $(y-3)=2m$.
$2x^3=2m(2m+6)$
$x^3=2m(m+3)$
Note that $x$ is even. So both of $m$ and $(m+3)$ are powers of 2.
Suppose, $m=2^a$ and $m+3=2^b$
$2^b-2^a=3$
It implies $a$ is 0. So $m=1$. Only solution for $x$ is $2$.
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
- Phlembac Adib Hasan
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Re: Perfect Square
$x$ even does not imply this, because you didn't prove $m$ or $m+3$ has no odd prime factors. Only you can guarantee exactly one of $m$ and $m+3$ is divisible by $4$.Fahim Shahriar wrote:Let $2x^3+9=y^2$
$2x^3=(y+3)(y-3)$
Surely $y$ is odd. Then $(y+3)$ and $(y-3)$ both will be even. Take $(y-3)=2m$.
$2x^3=2m(2m+6)$
$x^3=2m(m+3)$
Note that $x$ is even. So both of $m$ and $(m+3)$ are powers of 2.
After this step I split the problem into two cases and assumed $m=4k$. Suppose $x=2n$. So the equation becomes $n^3=k(4k+3)$. Now note that $\gcd(k,4k+3)|3$. Again two cases:
Assume $k,4k+3$ coprime. So they both should be perfect cube separately. So let $k=r^3, 4k+3=s^3\Longrightarrow 4r^3+3=s^3$. Now I am trying to solve this one.
Re: Perfect Square
Well, we should have that x=0 is a solution...
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Re: Perfect Square
Sorry for doing mistake in haste.
$d=gcd(y+3,y-3) = 2 or 6$
When d=2,
Let $(y+3)=2a$,$(y-3)=2b$ ; where $a,b$ are relatively prime. And $x=2k$
$16k^3 = 4ab$
$4k^3 = ab$
As $a,b$ are relatively prime. each divisor of k must enter one of them.
$k=uv$, $(a=4u^3,b=v^3)$ or $(a=u^3,b=4v^3)$
$4u^3-v^3=3$ or $u^3-4v^3=3$
Clearly two solutions are $u=v=1$ and $u=v=-1$
We get $x=2$ and $y= \pm 5$
Similarly, taking $d=6$ you will $x=6$ and $y= \pm 21$
Besides,factors indicates x=0.
$d=gcd(y+3,y-3) = 2 or 6$
When d=2,
Let $(y+3)=2a$,$(y-3)=2b$ ; where $a,b$ are relatively prime. And $x=2k$
$16k^3 = 4ab$
$4k^3 = ab$
As $a,b$ are relatively prime. each divisor of k must enter one of them.
$k=uv$, $(a=4u^3,b=v^3)$ or $(a=u^3,b=4v^3)$
$4u^3-v^3=3$ or $u^3-4v^3=3$
Clearly two solutions are $u=v=1$ and $u=v=-1$
We get $x=2$ and $y= \pm 5$
Similarly, taking $d=6$ you will $x=6$ and $y= \pm 21$
Besides,factors indicates x=0.
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Re: Perfect Square
One more thing I want to say-
It can have more solutions except those 3.
It can have more solutions except those 3.
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College