Problem:
 Score : 400 points    Problem Statement Give a pair of integers (A, B) such that A^5-B^5 = X .
It is guaranteed that there exists such a pair for the given integer X .      Constraints  1 \leq X \leq 10^9  X is an integer.  There exists a pair of integers (A, B) satisfying the condition in Problem Statement.         Input Input is given from Standard Input in the following format:  X       Output Print A and B , with space in between.
If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.  A  B         Sample Input 1 33      Sample Output 1 2 -1  For A=2 and B=-1 , A^5-B^5 = 33 .       Sample Input 2 1      Sample Output 2 0 -1

Solution (Python):
def main():
    X = int(input())
    for a in range(int(X**0.2)+1):
        for b in range(int(X**0.2)+1):
            if a**5 + b**5 == X:
                print(a, b*(-1))
                return
    for b in range(1, int(X**0.25)+1):
        temp = (X + b**5)**0.2
        if abs(temp - int(temp)) < 0.0001:
            print(int(temp), b)
            break
main()

Solution (Python):
x = int(input())
base = [0]
now = 1
while now ** 5 - base[-1] <= 10 ** 9:
  base.append(now ** 5)
  now += 1
for a,va in enumerate(base[1:],1):
  for b,vb in enumerate(base[:a]):
    if va - vb == x:
      print(a,b)
      break
    elif va + vb == x:
      print(a,-b)
      break

Solution (Python):
x=int(input())


targetList=[]
for i in range(-3000,3000):
    targetList.append(i**5)

for i in range(-3000,3000):
    for j in range(i,3000):
        if targetList[3000+j]-targetList[3000+i]==x:
            print(str(j)+" "+str(i))
            exit()

Solution (Python):
X = int(input())

div = []

sqrtX = int(X**0.5)+1
for i in range(1,sqrtX+1):
    if X % i == 0:
        div.append(i)
        div.append(X//i)

div = sorted(set(div))

for d in div:
    A = 0
    while True:
        B = A - d
        R = A**4 + A**3 * B + A**2 * B**2 + A* B**3 + B**4
        Y = (A-B) * R
        if Y == X:
            print(A,B)
            exit()

        B = d + A
        R = A**4 + A**3 * B + A**2 * B**2 + A* B**3 + B**4
        Y = (A-B) * R
        if Y == X:
            print(A,B)
            exit()
        if A > 1000:
            break
        A += 1

Solution (Python):
x = int(input())

for a in range(-1000, 1000):
  for b in range(-1000, 1000):
    if a ** 5 - b ** 5 == x:
      print(a, b)
      exit()

Solution (Python):


k=input()
flag=0
for i in range(-200,200):
    for j in range(i,200):
        xd=pow(j,5)-pow(i,5)
        if(xd==k):
            print j,i
            flag=1
            break
    if(flag):
        break

Solution (Python):
x=int(input())
for a in range(-200,200):
    for b in range(-200,200):
        if (a**5-b**5)==x:
            a_ans=a
            b_ans=b
            break

print(a_ans,b_ans)