<p>The secp256k1 curve is defined over x and y coordinates that are members of the finite field GF(2^256 - 2^32 - 977), or in other words, their operations hold only when considered modulo 115792089237316195423570985008687907853269984665640564039457584007908834671663.</p>

<p>55066263022277343669578718895168534326250603453777594175500187360389116729240^3 - (32670510020758816978083085130507043184471273380659243275938904335757337482424^2 + 7) is exactly 1442042049659660869506300006036683750029629333882594701370927246876626245108435922902327776681700708714008192087431130951749952236093997894375239788520937 times 115792089237316195423570985008687907853269984665640564039457584007908834671663.</p>

CC BY-SA 4.0
2021-02-27T05:51:20.203Z
@Harald1970 Yes.
Pieter Wuille
CC BY-SA 4.0
2020-06-11T17:28:38.257Z
Christ, don't you guys have MathJax on this site?
probably_someone
CC BY-SA 4.0
2020-06-11T17:41:32.460Z
@probably_someone: see https://bitcoin.meta.stackexchange.com/q/563/208
Pieter Wuille
CC BY-SA 4.0
2020-06-13T12:56:02.130Z
@PieterWuille Aha that explains it! Thank you. It has been a long time since I worked with modulo, so I am practicing a bit again. Is this a correct representation of graph g(x) = 2x + 1 (mod 4)? (graph in red) https://www.mediafire.com/convkey/c56a/tcux7iwkbkqf7razg.jpg
Harald1970
CC BY-SA 4.0
2020-06-11T00:31:13.680Z
2020-06-11T00:31:13.680Z
Pieter Wuille
<p>Finite field mathematics and elliptic curves don't use the &quot;normal operations&quot;. For example adding two finite field elements a and b isn't as simple as a + b. its actually
(a+b)%Prime where prime is the size of the finite field, this ensures the CLOSED property is meant. which says that if a is in the set and b is in the set than a + b is also in the set.
Same goes with multiplication which is what we have here.
a<em>b=(a</em>b)%Prime</p>
<p>so for gx and gy you can use this (gy^2) % (2^256 - 2^32 - 977) = ((gx^3) + 7) % (2^256 - 2^32 - 977). These are in fact are equal.</p>

CC BY-SA 4.0
2021-05-09T02:45:40.873Z
2021-05-09T02:45:40.873Z
mozahzah
<p>From what I understood are these the coordinates of basepoint G of Secp256K1 on the Elliptic Curve, in hexadecimal and decimal format.</p>

<p>Hexadecimal</p>

<p>Gx = (79BE667E F9DCBBAC 55A06295 CE870B07 029BFCDB 2DCE28D9 59F2815B 16F81798) </p>

<p>Gy = (483ADA77 26A3C465 5DA4FBFC 0E1108A8 FD17B448 A6855419 9C47D08F FB10D4B8)</p>

<p>Decimal</p>

<p>Gx = (55066263022277343669578718895168534326250603453777594175500187360389116729240)</p>

<p>Gy = (32670510020758816978083085130507043184471273380659243275938904335757337482424)</p>

<p>What I don't understand is that if I enter these values in the corresponding equation (y² = x³ + 7) as below, it seems that point G isn't on the curve at all. But it should, right? So what do I wrong? </p>

<pre>

(32670510020758816978083085130507043184471273380659243275938904335757337482424)²

is not equal to

(55066263022277343669578718895168534326250603453777594175500187360389116729240)³ + 7

</pre>

<secp256k1>
Why doesn't basepoint G of Secp256K1 seem to be on the Elliptic Curve?
Harald1970
# Why doesn't basepoint G of Secp256K1 seem to be on the Elliptic Curve?

(Asked by: username_0 on 2020-06-11)

From what I understood are these the coordinates of basepoint G of Secp256K1 on the Elliptic Curve, in hexadecimal and decimal format.

Hexadecimal

Gx = (79BE667E F9DCBBAC 55A06295 CE870B07 029BFCDB 2DCE28D9 59F2815B 16F81798)

Gy = (483ADA77 26A3C465 5DA4FBFC 0E1108A8 FD17B448 A6855419 9C47D08F FB10D4B8)

Decimal

Gx = (55066263022277343669578718895168534326250603453777594175500187360389116729240)

Gy = (32670510020758816978083085130507043184471273380659243275938904335757337482424)

What I don't understand is that if I enter these values in the corresponding equation (y² = x³ + 7) as below, it seems that point G isn't on the curve at all. But it should, right? So what do I wrong?

```

(32670510020758816978083085130507043184471273380659243275938904335757337482424)²

is not equal to

(55066263022277343669578718895168534326250603453777594175500187360389116729240)³ + 7

```

**Tags:** secp256k1


---

## 2 answers

### Answer by username_1 on 2020-06-11. Score: 11

The secp256k1 curve is defined over x and y coordinates that are members of the finite field GF(2^256 - 2^32 - 977), or in other words, their operations hold only when considered modulo 115792089237316195423570985008687907853269984665640564039457584007908834671663.

55066263022277343669578718895168534326250603453777594175500187360389116729240^3 - (32670510020758816978083085130507043184471273380659243275938904335757337482424^2 + 7) is exactly 1442042049659660869506300006036683750029629333882594701370927246876626245108435922902327776681700708714008192087431130951749952236093997894375239788520937 times 115792089237316195423570985008687907853269984665640564039457584007908834671663.

> **Comments:**
> **username_2**: Christ, don't you guys have MathJax on this site?  
> **username_1**: @username_2: see https://bitcoin.meta.stackexchange.com/q/563/208  
> **username_0**: @username_1 Aha that explains it! Thank you. It has been a long time since I worked with modulo, so I am practicing a bit again. Is this a correct representation of graph g(x) = 2x + 1 (mod 4)? (graph in red) https://www.mediafire.com/convkey/c56a/tcux7iwkbkqf7razg.jpg  
> **username_1**: @username_0 Yes.  

---

### Answer by username_3 on 2021-05-09. Score: -1

Finite field mathematics and elliptic curves don't use the "normal operations". For example adding two finite field elements a and b isn't as simple as a + b. its actually
(a+b)%Prime where prime is the size of the finite field, this ensures the CLOSED property is meant. which says that if a is in the set and b is in the set than a + b is also in the set.
Same goes with multiplication which is what we have here.
a*b=(a*b)%Prime

so for gx and gy you can use this (gy^2) % (2^256 - 2^32 - 977) = ((gx^3) + 7) % (2^256 - 2^32 - 977). These are in fact are equal.


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