Integer factorization $n = pq$ with additional knowledge of $\lfloor \sqrt{p} \rfloor \oplus \lfloor \sqrt{q}\rfloor$

vfenux

8/21/24, 1:48 PM

We know that we can factor integer $n = pq$ when we know that $p\oplus q$, where $\oplus$ means xor. If we know $\lfloor \sqrt{p} \rfloor \oplus \lfloor \sqrt{q}\rfloor$, can we factor $n$?

0 + 4

factoring

fgrieu

8/21/24, 3:12 PM

The question is unlikely to have any practical application to cryptography, but it's nevertheless fun. The answer might depend on $\log_2(n)$, and $-\log_2((p-q)/(p+q))$.

Myria

8/21/24, 8:49 PM

Do you have a link on how to factor if you have the XOR of the two factors? That sounds like a cool trick. Knowing $p + q$ also factors $pq$.

vfenux

8/22/24, 1:36 AM

@Myria, https://math.stackexchange.com/questions/2087588/integer-factorization-with-additional-knowledge-of-p-oplus-q

vfenux

8/22/24, 1:39 AM

@fgrieu, I guess we can find all possible $\lfloor\sqrt{p}\rfloor$ and $\lfloor\sqrt{q}\rfloor$ first, but I don't know what to do.

Elon Musk

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EN: Integer factorization $n = pq$ with additional knowledge of $\lfloor \sqrt{p} \rfloor \oplus \lfloor \sqrt{q}\rfloor$

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