Find all prime numbers $p$ for which $p^3-4p+9$ is a perfect square.
Problem
Source:
Tags: modular arithmetic, number theory proposed, number theory
31.08.2010 16:30
I wrote a solution that came to my mind at the moment, and I know that it is still missing some details, but I really don't want to write them down. It seems that one may continue taking consecutive moduli $4,8,16,\dots$ from the equation to resolve all cases, but there is always a lurking terror at the back of one's mind that this rabbit hole of taking consecutive moduli may be an endless pursuit, so the unwritten details are really left for you to do if you are not afraid.
31.08.2010 16:43
amparvardi wrote: Here is my solution : Let $p^3-4p+9 =a^2$ where $a$ is an integer. Now, we have $(p^2-1)^2=p^4 - 2p^2+1$ $\implies (p^2-1)^2-a^2=p^4 - 2p^2+1 - p^3 +4p -9$ $=p^4-p^3-2 p^2+4 p-8$ $=(p-2) (p+2) (p^2-p+2)$ $\implies (p^2-a-1)(p^2+a-1)=(p-2) (p+2) (p^2-p+2)$ We know that if $p>2$ then $\gcd(p-2,p+2,p^2-p+2)=1.$ This means $p^2 - a -1$ and $p^2 + a -1$ should divide just one of $p-2, p+2$ or $p^2-p+2.$ But $p^2-a-1$ doesn't divide $p-2, p+2$ or $p^2-p+2.$. for $p >2.$[This requires a lot of case works. I try to write them later.] Hence the unique solution $\boxed{p=2}$ crazyfehmy, do you have a neat solution for this ? My solution is not very nice too. By the way, $p=7$ and $p=11$ are also solutions since $7^3-4\cdot 7+9=18^2$ $11^3-4\cdot11 +9 = 36^2$
31.08.2010 16:47
Oops It seems my solution doesn't work. I should think more for a neat one.
31.08.2010 16:52
Probably not much help but: $p^{3} - 4p +9=n^{2}$ $p^{3} - 4p = n^{2}-9$ $(p-2)p(p+2) = (n-3)(n+3)$ (*) This is $0=0$ for $p=2$ and $n=3$. Presumably for the other possibility, or possibilities, we need to consider cases.
31.08.2010 16:59
Here is the official solution. We want to find the solutions of the equation $x^2=p^3-4p+9$ where $p$ is a prime number and $x$ is a non-negative integer. Since $x^2 \equiv 9 \pmod p,$ we have $x=kp\pm 3$ where $k$ is an integer. Then $(kp \pm 3)^2=p^3-4p+9 \: \Longrightarrow \: k^2p \pm 6k =p^2-4,$ and we obtain $ p \mid 6k \pm 4.$ If $p \neq 2,$ then $p \mid 3k \pm 2 \: \Longrightarrow \: p \leq 3k+2 \Longrightarrow \: \frac{p-2}{3} \leq k \: \Longrightarrow \: \frac{p^2-2p-9}{3} \leq pk-3 \leq x.$ $i) \:$ If $x \leq \frac{p^2}{4},$ then we have $\frac{p^2-2p-9}{3} \leq \frac{p^2}{4} \: \Longrightarrow \: p \leq 8+\frac{36}{p} \: \Longrightarrow \: p \leq 11.$ $ii) \: $ If, on the other hand, $x> \frac{p^2}{4},$ then $x^2=p^3-4p+9 \: \Longrightarrow \: \frac{p^4}{16}<p^3-4p+9 \: \Longrightarrow$ $p<16-\frac{16(4p-9)}{p^3} \: \Longrightarrow \: p \leq 13.$ Finally, for $p \leq 13, \: (p,x)=(2,3), \: (7,18),$ and $(11,36)$ are the only solutions.
31.08.2010 17:50
As in the official solution we have $x=kp\pm 3 $ and $ k^{2}p\pm 6k =p^{2}-4$ The discriminant of the last equation is $k^4 \pm 24k+16$ which must be a perfect square. But for $k>12$ $(k^2+1)^2>k^4+24k+16>(k^2)^2$ and $(k^2)^2>k^4-24k+16>(k^2-1)^2$. Now we need to find $p$ using the equation $ k^{2}p\pm 6k =p^{2}-4$ and $k={1,2,\ldots\,12}$.
29.06.2012 11:43
SAPOSTO wrote: As in the official solution we have $x=kp\pm 3 $ and $ k^{2}p\pm 6k =p^{2}-4$ The discriminant of the last equation is $k^4 \pm 24k+16$ which must be a perfect square. But for $k>12$ $(k^2+1)^2>k^4+24k+16>(k^2)^2$ and $(k^2)^2>k^4-24k+16>(k^2-1)^2$. Now we need to find $p$ using the equation $ k^{2}p\pm 6k =p^{2}-4$ and $k={1,2,\ldots\,12}$. and we can say $ k\equiv 0\pmod 3, $ Because $ k^{4}\pm 24k+16 $ which must be perfect square.
09.10.2014 15:58
I am sorry bumping old post butttt where can i have the official solution for all problems of turkey but in english languageee???
02.11.2014 22:35
There is NOT exist even turkish official ones (as far as i know).So you cant find them. is this solutions not enough for you?
19.06.2023 10:09
Amir Hossein wrote: Here is my solution : Let $p^3-4p+9 =a^2$ where $a$ is an integer. Now, we have $(p^2-1)^2=p^4 - 2p^2+1$ $\implies (p^2-1)^2-a^2=p^4 - 2p^2+1 - p^3 +4p -9$ $=p^4-p^3-2 p^2+4 p-8$ $=(p-2) (p+2) (p^2-p+2)$ $\implies (p^2-a-1)(p^2+a-1)=(p-2) (p+2) (p^2-p+2)$ We know that if $p>2$ then $\gcd(p-2,p+2,p^2-p+2)=1.$ This means $p^2 - a -1$ and $p^2 + a -1$ should divide just one of $p-2, p+2$ or $p^2-p+2.$ But $p^2-a-1$ doesn't divide $p-2, p+2$ or $p^2-p+2.$. for $p >2.$[This requires a lot of case works. I try to write them later.] Hence the unique solution $\boxed{p=2}$ crazyfehmy, do you have a neat solution for this ? Have you figured it out yet?
23.06.2023 06:05
mistakesinsolutions wrote: Have you figured it out yet? I've updated my old post and written a new solution, but it is still missing some details. I know that with enough trial and error, it is possible to find a way to take the equation modulo consecutive powers of $2$ to fill the details, but I'm intentionally refraining to do so.
23.06.2023 07:33
Amir Hossein wrote: mistakesinsolutions wrote: Have you figured it out yet? I've updated my old post and written a new solution, but it is still missing some details. I know that with enough trial and error, it is possible to find a way to take the equation modulo consecutive powers of $2$ to fill the details, but I'm intentionally refraining to do so.
When you said at $p=8k+3$ at the last part when you find$n$ and $y$, how can you find the solutions of this equation $y^2 = 256 n^3 + 528 n^2 + 359 n + 81.$?
23.06.2023 07:37
David_Kim_0202 wrote: When you said at $p=8k+3$ at the last part when you find$n$ and $y$, how can you find the solutions of this equation $y^2 = 256 n^3 + 528 n^2 + 359 n + 81.$? As I wrote in the second post on this thread, Amir Hossein wrote: I wrote a solution that came to my mind at the moment, and I know that it is still missing some details, but I really don't want to write them down. It seems that one may continue taking consecutive moduli $4,8,16,\dots$ from the equation to resolve all cases, but there is always a lurking terror at the back of one's mind that this rabbit hole of taking consecutive moduli may be an endless pursuit, so the unwritten details are really left for you to do if you are not afraid. The logical next step in this process (of taking consecutive powers of two as moduli) would be to take $y^2 = 256 n^3 + 528 n^2 + 359 n + 81$ modulo $16$. What would you get then?
12.09.2023 11:09
If $p \neq 2,$ then $p \mid 3k \pm 2 \: \Longrightarrow \: p \leq 3k+2 \Longrightarrow \: \frac{p-2}{3} \leq k \: \Longrightarrow \: \frac{p^2-2p-9}{3} \leq pk-3 \leq x.$ Question 1 : p ≤(3k+2) .........here why you didn't work on this case that p ≤(3k-2) plz help your brother...(help me) Ques 2: $x \leq \frac{p^2}{4},$ $x> \frac{p^2}{4},$ ... here I can't realize that how you known that ....plz help me how and by which case I can realize that $x> \frac{p^2}{4},$ or $x \leq \frac{p^2}{4},$ .
12.09.2023 11:36
If $p \neq 2,$ then $p \mid 3k \pm 2 \: \Longrightarrow \: p \leq 3k+2 \Longrightarrow \: \frac{p-2}{3} \leq k \: \Longrightarrow \: \frac{p^2-2p-9}{3} \leq pk-3 \leq x.$ Question 1 : p ≤(3k+2) .........here why you didn't work on this case that p ≤(3k-2) plz help your brother...(help me) Ques 2: $x \leq \frac{p^2}{4},$ $x> \frac{p^2}{4},$ ... here I can't realize that how you known that ....plz help me how and by which case I can realize that $x> \frac{p^2}{4},$ or $x \leq \frac{p^2}{4},$ .