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I have this task to prove that the factorization of number $n = p \cdot q$ (where $p$ and $q$ are prime) task is equivalent to finding square root module n.

I have found this lecture that explains the proof itself and I do not question it, it is the example they show of pages 19-29 that causes confusion. It is different from similar question Split $n$ into nontrivial factors via a nontrivial square-root of $1\!\pmod{\!n}$, because I do not need the proof, I just want to know how the numbers we square are chosen. I understand that the lecture may give some explanation, but to me, they look completely separate and do not explain it.

"Factor n = 299.

$96^2 \equiv 246 \mod n$". Where in this case 96 comes from??

It looks like all the numbers that have been chosen in the example are less than 100 and are not random.

https://hyperelliptic.org/tanja/teaching/crypto21/rsa-6.pdf

Any help would be greatly appreciated.

Charlotte
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  • Where does in this case $96$ comes from?? - did you see the next page? With $91^2\equiv 208\bmod n$, again for $n=299$? Where does the $91$ come from? Before is says: Pick random $1<a<n$ , let $c\equiv a^2 \bmod n$. – Dietrich Burde Jun 23 '24 at 15:49
  • @DietrichBurde, did yeah, so they just choose random numbers $a$ less than $n$ and for them solve $a^2 \equiv c \mod n$ and that's it? Because all the numbers are less than 100, it seemed like there was some kind of system in their actions – Charlotte Jun 23 '24 at 15:52
  • Yes, there is a "system", which is explained before the examples. – Dietrich Burde Jun 23 '24 at 15:58
  • @DietrichBurde maybe because it is not my first language, but their explanation made zero sense to me. But thanks for the input, I guess – Charlotte Jun 23 '24 at 16:01
  • Re: your edit, it is explained in the algorithm on the final page, i.e. they square random integers seeking some combination yielding a nontrivial square root over the factor base (cf. the approach of the quadratic sieve). – Bill Dubuque Jun 23 '24 at 17:23
  • Note that mod $299$ every integer $,n,$ is congruent to one of $,0,\pm1,\pm2,\ldots,\pm149,,$ so $,n^2,$ is congruent to one of $,0^2,1^2,\ldots, 149^2,,$ so we only need to square "random" naturals $< 149.\ $ That's why the chosen $,n,$ are small. Further, likely they chose numbers that make the computations easy by hand. $\ \ $ – Bill Dubuque Jun 23 '24 at 17:42
  • @BillDubuque, so they are actually random, but I should find such combination that will not give me $gcd (a, n) \not\equiv 1$, because i chose some random numbers that give me square at the end, but I stuck on this step. – Charlotte Jun 23 '24 at 19:17
  • What numbered step of the algorithm on the final page are you stuck on, and why? – Bill Dubuque Jun 23 '24 at 20:01

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