# RSA encrypting private keys blinding

I am studying RSA couldn't understand this question:

Alice encrypts the private factors of the modulus using her private key. In order to increase security, she multiplies them with a random integer $$k$$ she performs the following operations

$$cp = (k \cdot p)^e \bmod n \text{ and } cq = (k\cdot q)^e \bmod n,$$

n =
14522618414940330316720825703581818668232873090304101572437636625555724423016860272078630876437311699654574728861003760005749922965011995280733350140951078582251915088752445284367570052512244148190787248024210133865389997079370997487944695554576118063909791694130590309282475787773038629510027418889258775385528586923797001536550693747542175352795990447587607573178468163072638937166644494792819801190648017979291433924036979787320417318300097159771644354505341347447630385539964985486870484105658158920683576559610148645140369543464668029642924347163891332219762178122417025302960610955472998102936056418443726953009
e = 65537


Why this is not secure as anyone who obtains $$cp$$ or $$cq$$ can factor n.

Factor $$n$$ assuming

cp =
3873680249623467826367539364615622589341354860958676387978659380263546197984522526560815271059808890968051330871282433149509587198357995235098402597562394532942248440785823812744245775541457429705400656431200565210880345454073114392081573852975647397894188392554441729200872482027630124098341536309658821020887879754305515982200459718246770965592051443891902975413068576099241022023087113547866192282458339642152774673429640500380196758464923112739607056897968627591289881579460166394238775676402099831162820645802017259999853129954464949014713129189051795007415064128397315657343793939830810688657224114493477678984

• from my homework Nov 24, 2018 at 12:43

It is a simple Mathematics rule, given $$cp, e$$, and $$n$$ we have this relation:
$$(k \cdot p)^e \equiv cp \bmod n$$, then we can write this modular relation as;
$$k^e \cdot p^e = cp + l\cdot n$$, for some $$l \in \mathbb{Z}$$
$$p$$ divides $$n$$ that we already know, and from the equality, $$p$$ also divides the $$np$$