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As I know from the Elliptic Curve Diffie Hellman (ECDH), you can create a "Shared Key": https://asecuritysite.com/encryption/ecdh2

This will look like this:

dA × dB × G = Shared Key

QB = dB × G

QA = dA × G

But is it possible to get the sum of dA and dB? (Plusing the value of each private key?)

dA + dB × G = ???

How to get a "Shared Key" when adding each private key (plus)?

Izi Tors
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1 Answers1

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Yes, QA + QB = (dA + dB) × G, but this isn't used in ECDH as QA + QB can be calculated by third parties.

You can read about this property here.

MCCCS
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  • `QA + QB` can not be calculated by third parties? That is, the point G does not allow this to be done. Or how can `QA + QB` be calculated? Maybe in a different way? – Izi Tors Oct 25 '19 at 10:24
  • If you know the points QA and QB, what you should do is draw the curve, mark the points QA and QB, connect them with a line, find where the line intersects the curve (apart from QA and QB). The third intersection will be -(QA + QB). You can then reflect the point over the X axis to find the point QA + QB. You can read more about addition [here](https://fangpenlin.com/posts/2019/10/07/elliptic-curve-cryptography-explained/). (note that there are also some modulo operations) @IziTors – MCCCS Oct 25 '19 at 10:45
  • How on the graph of the curve can they get the sum of the total private keys (dA + dB) by calculating QA + QB? – Izi Tors Oct 25 '19 at 11:19
  • It’s assumed to be impossible to find dA + dB from QA + QB, but you can find (dA + dB) x G if you know QA and QB by adding them. If you know QA and QB or QA + QB you can’t find dA or dB or their sum. – MCCCS Oct 25 '19 at 12:03
  • And there are minus points QA - QB, and accordingly minus (dA - dB) x G ? – Izi Tors Nov 01 '19 at 12:27
  • @IziTors Yes. it's correct. – MCCCS Nov 01 '19 at 15:03
  • Replacing `y2` with `-y2` [is all you need to do](https://crypto.stackexchange.com/a/11336). @IziTors – MCCCS Jan 14 '20 at 18:13