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I have been given a public key ($y_1=x_1G$), but I don't yet know the private key ($x_1$). At a later time, I will get $x_1$ if some conditions are met.

Question:

Let's say I generate a new key pair $( y_2, x_2)$, both of which I will know.

Is it possible to add up these key pairs?

$ y = y_1 + x_2G$
$ x = x_1+x_2 $

Let's say (before I have $ x_1 $) I encrypt a text with $y$. And let's say I get $x_1$ at a later date. Can I decrypt my text using $x$ (by adding up the private keys)?

Let's use ECDSA key pairs. Sorry if my question is somehow deficient, but I just started to study cryptography. You would help me a lot by clarifying this question. Best, Stan

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Is it possible to add up these key pairs?

Yes. If $y_1=x_1\times G$ and $y_2=x_2\times G$, then $$y_1+y_2=y_1+(x_2\times G)=(x_1+x_2)\times G$$ and that's a public key matching private key $x=x_1+x_2$.

Let's say (before i have $ x_1 $) I encrypt a text with $y$. And let's say I get $x_1$ at a later date. Can I decrypt my text using $x$ (means by adding up the private keys)?

Yes, that follows from the above.

Note: there's no multiplication of key pairs as in the title, in the sense of an operation on keys yielding a new key (unless we note the group operation on the Elliptic Curve multiplicatively, which the question does not). With a group operation noted additively (as the question does), the only well-defined operation between public keys / group elements is addition, and we can only multiply these by a scalar (integer).

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  • $\begingroup$ You are great! Thank you very much for your explanation. With Multiplication, I understood now. Have a great day, Best Stan $\endgroup$ – StudentCryptoStan Jun 29 '18 at 14:59

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