RSA: Is there a way to digitally sign a message without knowing the private key? So, I am following an online course about cryptography these days, and I am stuck in an assignment. Here it is:
---------Known--------------
$m_1$: message 1
$m_1$ = 387
$s_1$: digital signature for $m_1$
$s_1$ = 104694095966994423550399840405679271451689287439740118881968798612714798360187905616965324945306116261186514828745033919423253353071142760352995900515279740753864505327750319034602409274084867637277266750899735986681083836544151016817569622466120342654469777762743360212628271549581658814852850038900877029382
$m_2$: message 2
$m_2$ = 2
$s_2$: digital signature for $m_2$
$s_2$ = 18269259493999292542402899855086766469838750310113238685472900147571691729574239379292239589580462883199555239659513821547589498977376834615709314449943085101697266417531578751311966354219681199183298006299399765358783274424349074040973733214578342738572625956971005052398172213596798751992841512724116639637
$m_3$: message 3
$m_3$ = 774
Signature scheme: Textbook
Public key
$e$ = 65537,
$n$ = 132177882185373774813945506243321607011510930684897434818595314234725602493934515403833460241072842788085178405842019124354553719616350676051289956113618487539608319422698056216887276531560386229271076862408823338669795520077783060068491144890490733649000321192437210365603856143989888494731654785043992278251
--------Task---------------
For a given $m_3$ create an $s_3$.
I tried to calculate $\phi(n)$ but because $n$ is too big I run out memory.
So, any ideas?
The comment by @Krystian is spot on. Textbook RSA is multiplicative so since $m_3=m_1 m_2,$ then $s_3=m_1^d m_2^d \pmod n=s_1 s_2 \pmod n.$
Magma tells me that $s_3$ should be

5483355855153602627619220309939701891434212768777923480007006960005208299649205
6581008264568421280116806103922182738655624242961539966642663712352455540402732
9086686980626004891264773131893337185873828146733471102601951184587947342940793
37057046673879975457092965542603806917235403924230185514038638412396010

if I haven't typed incorrectly.