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Suppose we have $a^2 \equiv b^2 \bmod 4R$ where $R$ is an integral domain. Under what conditions on $R$ can we conclude that $a \equiv b \bmod 2R$?

This would hold if $2 \in R$ is a prime or the product of distinct comaximal primes. This can fail when $R$ is not integrally closed: For $R = \mathbb{Z}[\sqrt{-3}]$ we have $(4+\sqrt{-3})^2 \equiv 1^2 \bmod 4R$ but $4 + \sqrt{-3} \not\equiv 1 \bmod 2R$.

Does this hold in a GCD domain? Does it hold, more generally, in an integrally closed domain?

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Denote $a-b=x$, then $a^2-b^2=x(x+2b)=4z$ for $z\in R$. Assuming that $R$ is integrally closed, we see that $(x/2)^2+b(x/2)-z=0$, so $x/2$ is an algebraic integer, thus $x/2\in R$.

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  • $\begingroup$ Does this work for cubes and higher powers, or is this one of those examples where squares are special? $\endgroup$
    – sqd
    Jul 14, 2020 at 21:12
  • $\begingroup$ Which claim for cubes do you mean? "If 9 divides $a^3-b^3$, then 3 divides $a-b$"? $\endgroup$
    – Fedor Petrov
    Jul 14, 2020 at 21:15
  • $\begingroup$ Yes. For an integrally closed integral domain $R$ do we have $a^3 \equiv b^3 \bmod 9R \implies a \equiv b \bmod 3R$? Do we have more generally for any $n > 2$ there exists an exponent $e$ such that $a^n \equiv b^n \bmod n^e R \implies a \equiv b \bmod nR$. $\endgroup$
    – sqd
    Jul 14, 2020 at 21:22
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    $\begingroup$ What about $a=1,b=w=e^{2\pi i/3}$ in $\mathbb{Z}[w]$? $\endgroup$
    – Fedor Petrov
    Jul 14, 2020 at 21:58
  • $\begingroup$ Yes, and this obviously generalizes: For $n > 2$ let $R = \mathbb{Z}[\theta]$ where $\theta = e^{2\pi i/n}$. Now for any $e \geq 1$ we have $1^n \equiv \theta^n \bmod n^e R$ but $1 \not\equiv \theta \bmod nR$. So this is special to squares. $\endgroup$
    – sqd
    Jul 14, 2020 at 22:12

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