Cryptography, Cybersecruity.
2¹⁹−1, g=3, and h=5 [here]: import random import sys n=pow(2,19)-1 def lipmaa(val): for i in range(0,100000): for j in range(0,1000): for k in range(0,100): for l in range(0,10): if (((i*i) + (j*j) + (k*k) + (l*l)) == val): return (i,j,k,l) return(0) x=999909 if (len(sys.argv)>1):