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2020-04-23 05:24:14 +0100 received badge  Student (source)
2020-04-21 13:04:38 +0100 commented answer Computing square root in IntegerModRing(2^n)

great! thanks not only for the reply but also for the explanation. I've accepted the answer, unfortunately cannot upvote it as well (not enough points). stay safe!

2020-04-21 13:03:49 +0100 received badge  Scholar (source)
2020-04-21 10:49:37 +0100 commented answer Computing square root in IntegerModRing(2^n)

thanks for the helpful quick reply. However, if the square root exists, shouldn't it be possible to compute it (exactly) still? In mathematica I can easily do so

a = Mod[1 + 2^(n - 1), 2^n];
root = PowerMod[a, 1/2, 2^n]
2020-04-21 09:37:40 +0100 asked a question Computing square root in IntegerModRing(2^n)

Hi, I'm trying to compute one square root in an IntegerModRing(2^n), but sage appears to fail to do so, am I doing something wrong?

PoC:

sage: n = 216
sage: FF = IntegerModRing(2^n)
sage: FF(1 + 2^(n - 1)).sqrt()

Thanks!

2020-04-21 09:37:40 +0100 asked a question alkjdspaodsips

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