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I can't find the proof for this. I know that $\mathbb{Z}[i]/(p)$ is an integral domain, so I tried to assume that there is a solution for $x \in \mathbb{Z}[i]$ and reach a contradiction of that fact, or the fact that $p$ is prime, but I wasn't able to do it. Can you help me? Thanks!

Arturo Magidin
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Brian Mac Guire
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    Hint: $(x+i)(x-i)=x^2+1 \in (p)$, a prime ideal – lhf Dec 15 '21 at 16:42
  • Thanks! I tried that. In that case, since $(p)$ is prime ideal, then I can assume that $x+i \in (p)$, so $x+i = kp$ for some $k \in \mathbb{Z}[i]$. Then, I can write that as $a + (b+1)i = cp + dpi$ where $x = a+bi$ and $k = c+di$. So $p|a$ and $p|b+1$. I can't go further then. Can you help me? Thanks! @lhf – Brian Mac Guire Dec 15 '21 at 17:40
  • A rational prime $p>2$ stays prime in $\Bbb Z[i]$ iff $p\equiv 3\bmod 4$, and then $x^2\equiv -1\bmod p$ has no solution, see here. – Dietrich Burde Dec 15 '21 at 17:59
  • In the other direction, if $x^2 \equiv -1 \pmod p$ has a solution, then we may construct $a^2 + b^2 = p.$ For example, there are algorithms. – Will Jagy Dec 15 '21 at 18:10
  • S. Wagon: "On page 144 of this issue of the MONTHLY Don Zagier presents an extremely short, elegant, and elementary proof of the classic result that any prime p congruent to 1 (mod 4) is a sum of two squares. The problem of finding a representation of p as a2 + /32 often arises in computational number theory; for example, it is a key step in factoring Gaussian integers into prime Gaussian integers. Fortunately, this problem can be solved by a very fast and easy-to-program algorithm-it is not widely known, although it uses nothing beyond undergraduate mathematics. " – Will Jagy Dec 15 '21 at 18:15
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    You probably mean — in the title and the question — that $x^2\equiv -1\pmod p$ has no solution in $\mathbb{Z}$, not $\mathbb{Z}[i]$; in the latter it has the 'standard' solution $x=i$. – Steven Stadnicki Dec 15 '21 at 18:16

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