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Does $x^2 = -1 \mod 5987$ have an integer solution?

Attempt: I know that $p=5987$ is prime, so $\phi(p)=p-1=5986$, so we need to know whether there exists an integer $k$ such that $x^2 = \phi(p) + kp = 5986 + 5987k$ has a solution. Any help would be great. Thanks!

Rócherz
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Jeffrey
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    What is the structure of the group of units? How many elements are in the group of units? Can you combine these facts to determine when the group of units has an element of order four? – anon Feb 28 '15 at 21:38
  • For which primes is $-1$ a quadratic residue? Is $5987$ of this form? – rogerl Feb 28 '15 at 21:40
  • The question has been handled countlessly many times on our site. $-1$ is a quadratic residue modulo an odd prime $p$, iff $p\equiv1\pmod4$. I don't know what is the best original question. The one that came up in my google search is the one I linked to. It has the benefit of giving an explicit square root. But this is not necessary. If somebody finds a better original, ping me or another moderator. – Jyrki Lahtonen Feb 28 '15 at 22:16
  • Another related question. – Jyrki Lahtonen Feb 28 '15 at 22:26
  • @Jyrki The question linked in the dupe closure is not a dupe, but the link in your prior comment is. – Bill Dubuque Feb 28 '15 at 22:41

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Hint $\ {\rm mod}\ p\!:\ x^2\equiv -1\,\Rightarrow\, x^4\equiv 1\,$ so $\,x\,$ has order $\,4.\,$ Thus $\,x^{p-1}\equiv 1\,\Rightarrow\, 4\mid p-1$

Remark $\ $ The order inference is a very special case of the general result that $\,a\,$ has order $\,n\,$ if $\,a^n = 1\,$ but $\,a^{n/p} \ne 1\,$ for every prime $\,p\mid n.\,$ The proof is easy: if the order $\,k\,$ is proper divisor of $\,n\,$ then $\,k\,$ arises by deleting at least one prime $\,p\,$ from the prime factorization of $\,n,\,$ so $\,k\mid n/p,\,$ say $\, jk = n/p,\,$ so $\,a^{n/p} \equiv (a^k)^j\equiv 1^j\equiv 1,\,$ contra hypothesis.

Bill Dubuque
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