Bangladesh IMO TST 1: 2011/ NT-2 (P 1)

[Moon]

Problem 1:
Find all prime numbers $p$ for which $p^3-4p+9$ is a perfect square.

[Masum]

Note that by Fermat's Little Theorem, $p^3-4p+9\equiv p-p\equiv 0\pmod 3$.Since it is a square,it must be divisible by $9,$so let
$p^3-4p+9=9x^2\Longrightarrow p(p^2-4)=9(x+1)(x-1)$.We have four cases:
Case 1:
$p=2$ which is trivially a solution.So let's concentrate on $p$ odd.
Before proceeding on,we first prove a lemma.
Lemma:
$v_p(x+1)=1$ or $v_p(x-1)=1$
Proof:
Let $x+1=p^kl$ with $gcd(p,l)=1$
Then $p(p^2-4)=9p^kl(p^kl-2)\Longrightarrow p^2-4=9p^{k-1}l(p^kl-2)$
Now if $k>1,$taking $\pmod p,$we get that $p|p^2-4\Longrightarrow p|4\Longrightarrow p=2$,contradiction.
Get back to the original problem.
Case 2: $p|9$ or $p=3$
Then it can be easilly verified that it is not a solution
Case 3: $p|x+1$,let $x+1=pk,k\ge 1$
Then $9k(pk-2)=p^2-4\Longrightarrow p^2=9pk^2-18k+4\Longrightarrow p|9k-2\Longrightarrow 9k\ge p+2$
We have $9k(pk-2)\ge (p+2)(pk-2)$ or $(p+2)(p-2)\ge (p+2)(pk-2)\Longrightarrow k\le 1$ that is,$k=1$ and $p=7$
Case 4: $x-1=pk$
Then $9k(pk+2)=p^2-4\Longrightarrow p|9k+2$ and in a similar approach $k=1,p=11$

Thus $p=\boxed {2,7,11}$ are the only solutions.

[TIUrmi]

Then $p(p^2-4)=9pk(pk-2)\Longrightarrow p^2-4=9k(pk-2)\Longrightarrow p^2+14=9pk^2$

Then $p(p^2-4)=9pk(pk+2)\Longrightarrow p^2=9pk^2+22$

$p^2-4=9k(pk-2)$ = $p^2+18k-4=9pk^2$ not $p^2+14=9pk^2$ or did I miss something?
Final solution is correct though!

[Masum]

Sorry for the careless mistake.I am editing,but still it would remain correct

[TIUrmi]

Actually it's still not correct. $18k-4 \not= 14k$ right? and $-18k -4\not=22k$

[Masum]

See now
I don't understand why I am so careless now-a-days.

[TIUrmi]

Don't worry! You are not alone. None other can beat me in carelessness..

[Masum]

but,my mistake is too hard to consider

[TIUrmi]

Then definitely you don't know my mistakes..