Find the least odd prime factor of $155^8+1$

Question

Find the least odd prime factor of $155^8+1$. How do I do this without using Wolfram Alpha or something?

Answer 1

$155^8\equiv-1\pmod p$ means $155^{16}\equiv1\pmod p.$

We know $155^{p-1}\equiv1\pmod p$.

That means $16|p-1$,

so candidates for $p$ are $17$, $97$, $\color{red}{113}$, ...

Answer 2

Following is a Python program that uses the Tonelli-Shanks algorithm;
the code for the algorithm is here at codereview.stackexchange.com.

The output lines

print(p, alg_list_prime_mod_sqrt_of_minus_one, res)

shows for each prime $p$ a list of solutions to $x^2 \equiv -1 \pmod p$ while $\text{res } \equiv {155}^4 \pmod p$.
When $\text{res}$ is a square root of $-1$ $\text{modulo-}p$ and in the list the program terminates.

Python program:

from math import sqrt; from itertools import count, islice

def isPrime(n):
    return n > 1 and all(n%i for i in islice(count(2), int(sqrt(n)-1)))


def legendre_symbol(a, p):
    """
    Legendre symbol
    Define if a is a quadratic residue modulo odd prime
    http://en.wikipedia.org/wiki/Legendre_symbol
    """
    ls = pow(a, (p - 1)//2, p)
    if ls == p - 1:
        return -1
    return ls

def prime_mod_sqrt(a, p):
    """
    Square root modulo prime number
    Solve the equation
        x^2 = a mod p
    and return list of x solution
    http://en.wikipedia.org/wiki/Tonelli-Shanks_algorithm
    """
    a %= p

    # Simple case
    if a == 0:
        return [0]
    if p == 2:
        return [a]

    # Check solution existence on odd prime
    if legendre_symbol(a, p) != 1:
        return []

    # Simple case
    if p % 4 == 3:
        x = pow(a, (p + 1)//4, p)
        return [x, p-x]

    # Factor p-1 on the form q * 2^s (with Q odd)
    q, s = p - 1, 0
    while q % 2 == 0:
        s += 1
        q //= 2

    # Select a z which is a quadratic non resudue modulo p
    z = 1
    while legendre_symbol(z, p) != -1:
        z += 1
    c = pow(z, q, p)

    # Search for a solution
    x = pow(a, (q + 1)//2, p)
    t = pow(a, q, p)
    m = s
    while t != 1:
        # Find the lowest i such that t^(2^i) = 1
        i, e = 0, 2
        for i in range(1, m):
            if pow(t, e, p) == 1:
                break
            e *= 2

        # Update next value to iterate
        b = pow(c, 2**(m - i - 1), p)
        x = (x * b) % p
        t = (t * b * b) % p
        c = (b * b) % p
        m = i

    return [x, p-x]

for p in range(2,555):
    if isPrime(p):
        mult = (155 % p)
        res = 1
        for dum in range(0,4):
            res = (res * mult) % p
        alg_list_prime_mod_sqrt_of_minus_one = prime_mod_sqrt(-1, p)
        print(p, alg_list_prime_mod_sqrt_of_minus_one, res)
        if res in alg_list_prime_mod_sqrt_of_minus_one:
            if p % 2 == 1:
                print('The number', p , 'is the smallest odd prime factor for 155^4 + 1.')
                break

$\text{*** OUTPUT ***}$

2 [1] 1
3 [] 1
5 [3, 2] 0
7 [] 1
11 [] 1
13 [8, 5] 1
17 [4, 13] 16
19 [] 5
23 [] 8
29 [12, 17] 24
31 [] 0
37 [6, 31] 33
41 [32, 9] 1
43 [] 15
47 [] 17
53 [23, 30] 44
59 [] 26
61 [11, 50] 20
67 [] 47
71 [] 19
73 [46, 27] 64
79 [] 2
83 [] 33
89 [34, 55] 25
97 [22, 75] 88
101 [91, 10] 68
103 [] 58
107 [] 39
109 [33, 76] 63
113 [98, 15] 15
The number 113 is the smallest odd prime factor for 155^4 + 1.