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Find all primes $p$ such that $ p^3-4p+9 $ is a perfect square.

I tried a few different values for $p$, namely $2,3,5,7,$ and $11$. The prime $p =2,7,11$ all worked but $p =13$ didn't so it makes me wonder. How can I find all primes such that it is a perfect square?

EDIT: it turns out this is problem P25 in post number #63 at http://artofproblemsolving.com/community/c3h1171106p5665470

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    $\begingroup$ If $p^3-4p+9 = n^2$ then $p^3-4p = n^2-9$, i.e. $(p-2)p(p+2) = (n-3)(n+3)$. I'm not sure if this helps, but it might be a good place to start. $\endgroup$
    – JimmyK4542
    Dec 17, 2015 at 21:53
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    $\begingroup$ A small beginning: $p^3-4p+9$ is always divisible by three. But it is divisible by nine only when $p\equiv\pm2\pmod9$. This rules out primes not satisfying that congruence. $\endgroup$ Dec 17, 2015 at 21:54
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    $\begingroup$ Also sprach Mathematica: Those three are the only primes that work among the 10000 smallest primes. $\endgroup$ Dec 17, 2015 at 22:02
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    $\begingroup$ @Myself , I like to check these things without restricting the variable to be prime. Sometimes that condition is a deliberate red herring to disguise the real techniques for dealing with the problem. $\endgroup$
    – Will Jagy
    Dec 17, 2015 at 22:08
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    $\begingroup$ A related question: it seems that $m^4 + 24m +16$ is a square only when $m=3$ (I checked up to $m = 50000000$). Arrive at that by solving for $p$ in the equation $p^3 - 4p + 9 = (mp + 3)^2$ (there is also the case of $mp-3$ I suppose. $\endgroup$
    – tkr
    Dec 17, 2015 at 22:57

2 Answers 2

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If $p^3-4p+9=n^2$ then $n^2\equiv 9 \pmod{p}$ and $\pm n \equiv 3 \pmod{p}$ since $n^2-9\equiv 0$ can only have two roots modulo a prime.

By choosing the appropriate sign we can write $$ \begin{align} n^2=(-n)^2 &= (kp+3)^2 \\ p^3-4p+9 &= k^2p^2+6kp+9 \\ 0 &= p^2-k^2p-(6k+4) \\ p & = \frac{k^2\pm \sqrt{k^4+24k+16}}{2} \end{align} $$ where we used $p\ne 0$ since it's a prime. This can only have an integer solution for $p$ when $k^4+24k+16$ is a square (as @tkr suggested in the comments). But $$ \begin{array}{ll} (k^2-1)^2 = k^4-2k^2+1 < k^4+24k+16<(k^2)^2 & \text{if }k<-11 \\ (k^2)^2 < k^4+24k+16< k^4+2k^2+1 =(k^2+1)^2 & \text{if }k>12 \end{array} $$ so if $k<-11$ or $k>12$ then $k^4+24k+16$ lies between two squares and hence is not a square. This leaves 24 values to check, which you could do by hand. It leads to solutions when $k=-3,0,3$ which give integral values for $p$ of $-2,2,7,11$, of which only $2,7,11$ would normally be considered prime.

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  • $\begingroup$ I got to your $ p^2-k^2p-(6k+4) = 0$ and solved for $k$ with the quadratic formula, but did not think to solve for $p$ with the quadratic formula, which is what does the job. $\endgroup$
    – Will Jagy
    Dec 18, 2015 at 18:32
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    $\begingroup$ Anyway, well done, and you also posting on AOPS is the right outcome. $\endgroup$
    – Will Jagy
    Dec 18, 2015 at 19:52
  • $\begingroup$ Here is a less involved solution on AoPS. $\endgroup$
    – Toby Mak
    Dec 19, 2020 at 8:04
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Zander's solution is nice, but there might be a way to reduce the number of cases to check. when $k\leq -6$ or $k\geq 7$
$(k^2-2)^2<k^4+24k+16<(k^2+2)^2$
but $k^4+24k+16=k^4,(k^2-1)^2,(k^2+1)^2$ don't have solutions. (compare parity for $(k^2-1)^2$ and $(k^2+1)^2$)
so $-5\leq k\leq 6$
also observe 3|k. if not, $k^4+24k+16\equiv 2\pmod3$ , which is not a square.
now $k\in\{-3,0,3,6\}$ , of which 3 out of the 4 cases yield solutions.

question is from turkey NMO 2009, 2nd round, q1: https://artofproblemsolving.com/community/c6h364542

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