Creating a nothing-up-my-sleeve public key

Question

I am trying to create a public key whose x starts with "34" and is followed by the minimum number n such that the concatenation of "34" and n is the x of a valid point (x, y) on the elliptic curve secp256k1. Submit the concatenation of 04, x and y in hexadecimal form. This is a valid bitcoin public key, the corresponding secret key of which is not known by anyone.

I was told that the minimum number is 0.

My understanding is that the answer is supposed to be like this: '0434' + hex(n) + y_value. I am not sure how to approach this.

Is it true that x-coordinate is supposed to be 32 bytes and y coordinate 32 bytes too?

Is it true that each coordinate is represented in hex mode of 64 chars?

Is the public key represented by 130 hex chars? I have tried some online tools: e.g. https://iancoleman.io/bitcoin-key-compression/ and some python code but I always get a wrong answer.

X = '0x3400000000000000000000000000000000000000000000000000000000000000'
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
print("x coordinate = %s " % X)
x = int(X,16)
print(x)
ysquared = ((x*x*x+7) % p)
print("ysquared= %s " % ysquared)
y = pow(ysquared, (p+1)//4, p)
print("y1 = %s " % hex(y))
print("y2 = %s " % hex((y * -1 % p)))
print("-----------------")
print('04' + X[2:] + hex(y)[2:])
print('04' + X[2:] + hex((y * -1 % p))[2:])

Answer 1

I was told that the minimum number is 0.

That is correct. 0x3400000000000000000000000000000000000000000000000000000000000000 is a valid x-coordinate on secp256k1.

My understanding is that the answer is supposed to be like this: '0434' + hex(n) + y_value. I am not sure how to approach this

That is also correct. However, you will have two y-coordinates. One even and one odd and hence two uncompressed public keys. That is the reason why we append a prefix to the x-coordinate to tell us whether the y-coordinate associated with that x-coordinate is even or odd.

Is it true that x-coordinate is supposed to be 32 bytes and y coordinate 32 bytes too?

That is the maximum size. It can be below that as well.

Is it true that each coordinate is represented in hex mode of 64 chars?

Related to to above. One byte representation needs two hexadecimal characters.

Python code

I'll use the below one.

p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
def modular_sqrt(n, p):
    ''' Find a quadratic residue (mod p) of 'n'. For Bitcoin p is an odd prime, that satisfies
        the identity p % 4 == 3. As a result, this is a simple solution where the modular square
        root of y^2 is : y = pow(y^2, (p + 1) // 4, p).
        If p did not satisfy that identity we would have had to use the Tonelli-Shank's algorithm
    '''
    return pow(n, (p + 1) // 4, p)

def y_coordinate_from_x(x_coordinate_with_prefix):
    public_key_x_coordinate = int(x_coordinate_with_prefix[2:], 16)
    public_key_y_coordinate_sign = int(x_coordinate_with_prefix[:2])
    y_squared = (public_key_x_coordinate**3 + 7) % int(p)
    public_key_y_coordinate = modular_sqrt(y_squared, int(p))

    # check the sign (odd or even) and match that to the sign provided by x-coordinate
    sign_check = 0 if public_key_y_coordinate_sign == 2 else 1
    if int(str(public_key_y_coordinate)[-1:]) % 2 != sign_check:
        public_key_y_coordinate = int(p) - public_key_y_coordinate

    return '%064x'%public_key_y_coordinate

x = '3400000000000000000000000000000000000000000000000000000000000000'
y1 = y_coordinate_from_x('02' + x)
y2 = y_coordinate_from_x('03' + x)

so your uncompressed public keys are '04'+x+y1 and '04'+x+y2