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Is it possible to use Chinese Remainder Theorem to reconstruct the GCD of two integers from several GCDs of their modular representations (i.e. residues modulo pair-wise coprime integers)? For example:

$$GCD(18, 24) = 6$$

$$ GCD(18 \bmod 7, 24 \bmod 7) = GCD(4, 3) = 1\\ GCD(18 \bmod 11, 24 \bmod 11) = GCD(7, 2) = 1\\ CRT(\langle 1, 7 \rangle, \langle 1, 11\rangle) = 1 $$

I'm asking because for polynomials something similar works: first compute several pairs of polynomials modulo prime and reconstruct the original GCD from the images via CRT. What is different compared to integers that here this works and for integers it (seemingly) does not?

Related:

GCD computation in Modular Residue Number System

Edit: For completeness, here is an example that does work:

$$GCD(30, 45) = 15$$

$$ GCD(30 \bmod 7, 45 \bmod 7) = GCD(2, 3) = 1\\ GCD(30 \bmod 13, 45 \bmod 13) = GCD(4, 6) = 2\\ CRT(\langle 1, 7 \rangle, \langle 2, 13 \rangle) = 15 $$

vvg
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Ecir Hana
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1 Answers1

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This is not a robust answer, but provides a heuristic reasoning for why the approach proposed in the question may not work.

For a suitably arbitrary prime $p$, a pair of residues $x,y \pmod{p}$ are likely to be coprime i.e., their GCD is likely to be $1$ with more than $0.5$ probability.

For example, there are $13$ coprime pairs modulo $5$ (therefore, coprimality probability = $0.52$) and $93$ pairs modulo $13$ (coprimality probability = $0.55$) and the limit for the probability is $6 \over \pi^2$ as $p \rightarrow \infty$. This is approximately $0.61$. (See this question and answer for more info).

If the prime doesn't divide the integers $x,y$ then we have to ignore the residue pairs $(0, y)$ and $(x, 0)$ and then the effective coprimality probability would go up even higher.

So, using even a probabilistic argument, we are likely to see $1$ as the GCD of the residues quite often even when the numbers themselves are not coprime.

Using this and the pigeon-hole principle, we should be finding the actual GCD only by chance (and very low at that).

vvg
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