factoring $n=pq$ knowing $p+q$ in polynomial time

Question

Lets $n=pq$ with $p$ and $q$ primes

If we know the result of $p+q$, can we factorize $n$ in polynomial time ?

Best regards.

Your thoughts please... (since this is not a 'do my homework for free' service). — barak manos, Jan 02 '17 at 16:36
Hope it was homework, but school is far far away..., I mean does knowing the sum of the two factors can help to factorize a number — BenLaz, Jan 02 '17 at 16:40

Asinomás · Accepted Answer · 2017-01-02T16:45:06.320

3

Suppose $n=pq$ and $p+q=k$, then $p$ and $q$ are the roots of the polynomial $x^2-(p+q)x+pq$ and so they are equal to $\frac{p+q\pm \sqrt{p+q)^2-4pq}}{2}$

edited Jan 02 '17 at 16:45

answered Jan 02 '17 at 16:38

Asinomás

105,651

right, thanks.${}{}{}{}{}$ – Asinomás Jan 02 '17 at 16:44
Great observation, but for completeness' sake, maybe including a conclusion about the polynomial time would be nice? – sTertooy Jan 02 '17 at 16:46
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@SteamyRoot: the time will be polylog, at most $\log^2n$. – Jan 02 '17 at 16:48
it is $\mathcal O(\log(n))$ as we need to find the square root of a number of size approximately $4n^2$. – Asinomás Jan 02 '17 at 16:49
@JorgeFernándezHidalgo: don't you mean like $O(\log n\log\log n\log\log\log n)$ ? – Jan 02 '17 at 16:52
@YvesDaoust why is there an $n$ all by itself? – Asinomás Jan 02 '17 at 16:55
@JorgeFernándezHidalgo: I fixed the typo. – Jan 02 '17 at 16:55
@YvesDaoust I guess that makes sense ;) – Asinomás Jan 02 '17 at 16:57

factoring $n=pq$ knowing $p+q$ in polynomial time

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