# ポラード・ロー素因数分解法
# https://qiita.com/t_fuki/items/7cd50de54d3c5d063b4a#%E3%83%9D%E3%83%A9%E3%83%BC%E3%83%89%E3%83%AD%E3%83%BC%E7%B4%A0%E5%9B%A0%E6%95%B0%E5%88%86%E8%A7%A3%E6%B3%95%E3%81%AE%E3%82%A2%E3%83%AB%E3%82%B4%E3%83%AA%E3%82%BA%E3%83%A0

def gcd(a, b):
    while a:
        a, b = b%a, a
    return b


def is_prime(n):
    if n == 2:
        return 1
    if n == 1 or n%2 == 0:
        return 0

    m = n - 1
    lsb = m & -m
    s = lsb.bit_length()-1
    d = m // lsb

    test_numbers = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]

    for a in test_numbers:
        if a == n:
            continue
        x = pow(a,d,n)
        r = 0
        if x == 1:
            continue
        while x != m:
            x = pow(x,2,n)
            r += 1
            if x == 1 or r == s:
                return 0
    return 1


def find_prime_factor(n):
    if n%2 == 0:
        return 2

    m = int(n**0.125)+1

    for c in range(1,n):
        f = lambda a: (pow(a,2,n)+c)%n
        y = 0
        g = q = r = 1
        k = 0
        while g == 1:
            x = y
            while k < 3*r//4:
                y = f(y)
                k += 1
            while k < r and g == 1:
                ys = y
                for _ in range(min(m, r-k)):
                    y = f(y)
                    q = q*abs(x-y)%n
                g = gcd(q,n)
                k += m
            k = r
            r *= 2
        if g == n:
            g = 1
            y = ys
            while g == 1:
                y = f(y)
                g = gcd(abs(x-y),n)
        if g == n:
            continue
        if is_prime(g):
            return g
        elif is_prime(n//g):
            return n//g
        else:
            return find_prime_factor(g)


def factorize(n):
    res = {}
    while not is_prime(n) and n > 1:  # nが合成数である間nの素因数の探索を繰り返す
        p = find_prime_factor(n)
        s = 0
        while n%p == 0:  # nが素因数pで割れる間割り続け、出力に追加
            n //= p
            s += 1
        res[p] = s
    if n > 1:  # n>1であればnは素数なので出力に追加
        res[n] = 1
    return res

# 高速約数列挙
def divisors(num):
    factors = factorize(num)
    divs = [1]
    for p in factors:
        e = factors[p]
        if e > 0:
            k = len(divs) #それまでの素因数積、つまり約数、の数
            for i in range(e*k):
                divs.append(divs[-k]*p) 
                #なぜans[-k]なのか、どんどんappendするので[-k]で前の約数にかけていく
    return divs

# TLEしたので高速約数列挙を作った
# 因数分解できる形に持ち込むことが必要
# 1/(3x) + 1/(3y) = 1/2の解き方と同じだった
# 変形してQM+QN=PMN
# PPMN - PQM - PQN = 0
# (PM-Q)*(PN-Q) - QQ=0
# (PM-Q)*(PN-Q) = Q**2

from math import gcd
P, Q = map(int, input().split())
g = gcd(P, Q)
P = P//g
Q = Q//g
    
divs = divisors(Q**2)
ans_set = set()
#print('P', P, 'Q', Q, 'divs', divs)
for d in divs:
    d2 = (Q**2)//d
    if (Q+d)%P == 0 and (Q+d2)%P == 0:
        m = (Q+d)//P
        n = (Q+d2)//P
        ans_set.add((n, m))
        #print(d, d2, n, m)
ans_list = sorted(list(ans_set))
print(len(ans_list))
for n, m in ans_list:
    print(n, m)