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How much primes p exist so that $p^3-4p+9$ is a perfect square?

Of course we have p = 2, which is basically 9. I have no idea what to do now. I am trying to write it as a product a long time, but i am concluding that it is impossible if we just want integers in the product. Anyway, i don't know another method to seek the others primes.

Gabriela Da Silva
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    A necessary condition for odd primes $p$ is $p \equiv 3 \pmod{4}$ (this follows from the fact that perfect squares are congruent to $0$ or $1$ modulo $4$). It seems $7$ and $11$ both work. The next few primes congruent to $3$ modulo $4$ don't seem to work however. – jl00 Dec 19 '20 at 07:37
  • This problem has been posted multiple times on AoPS. Please use Approach0 to search for duplicates. – Toby Mak Dec 19 '20 at 08:01
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    Thank you all. Really beautiful solution in this topic, @TobyMak. What is this AOPS? – Gabriela Da Silva Dec 19 '20 at 08:09
  • Art of Problem Solving: a forum for discussing contest maths questions and Olympiad problems. You should check it out if you're interested. – Toby Mak Dec 19 '20 at 08:17

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