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

<p>You can read about this property <a href="https://crypto.stackexchange.com/a/53980/">here</a>.</p>

CC BY-SA 4.0
2019-10-25T10:24:25.593Z
`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
CC BY-SA 4.0
2019-10-25T10:45:12.803Z
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
CC BY-SA 4.0
2019-10-25T11:19:06.393Z
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
CC BY-SA 4.0
2019-10-25T12:03:18.450Z
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
CC BY-SA 4.0
2019-11-01T12:27:02.017Z
And there are minus points QA - QB, and accordingly minus (dA - dB) x G ?
Izi Tors
CC BY-SA 4.0
2019-11-01T15:03:41.363Z
@IziTors Yes. it's correct.
MCCCS
CC BY-SA 4.0
2020-01-14T18:13:37.753Z
Replacing `y2` with `-y2` [is all you need to do](https://crypto.stackexchange.com/a/11336). @IziTors
MCCCS
CC BY-SA 4.0
2019-10-25T10:01:44.023Z
2019-10-25T10:09:18.890Z
2019-10-25T10:09:18.890Z
MCCCS
<p>As I know from the Elliptic Curve Diffie Hellman (ECDH), you can create a "Shared Key": <a href="https://asecuritysite.com/encryption/ecdh2" rel="nofollow noreferrer">https://asecuritysite.com/encryption/ecdh2</a></p>

<p>This will look like this:</p>

<pre><code>dA × dB × G = Shared Key

QB = dB × G

QA = dA × G
</code></pre>

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

<pre><code>dA + dB × G = ???
</code></pre>

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

<private-key><public-key><ecdsa><secp256k1><shared>
Is it possible to add two secp256k1 points to get a "shared key"?
Izi Tors
# Is it possible to add two secp256k1 points to get a "shared key"?

(Asked by: username_0 on 2019-10-25)

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)?

**Tags:** private-key, public-key, ecdsa, secp256k1, shared


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## 1 answers

### Answer by username_1 on 2019-10-25. Score: 1

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](https://crypto.stackexchange.com/a/53980/).

> **Comments:**
> **username_0**: `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?  
> **username_1**: 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  
> **username_0**: How on the graph of the curve can they get the sum of the total private keys (dA + dB) by calculating QA + QB?  
> **username_1**: 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.  
> **username_0**: And there are minus points QA - QB, and accordingly minus (dA - dB) x G ?  
> **username_1**: @IziTors Yes. it's correct.  
> **username_1**: Replacing `y2` with `-y2` [is all you need to do](https://crypto.stackexchange.com/a/11336). @IziTors  

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