This is RSA with a small public exponent. It can be broken by taking the third root of the ciphertext:--------------------------------------------------------import gmpyfrom Crypto.Util.number import long_to_bytesn = 92164540447138944597127069158431585971338721360079328713704210939368383094265948407248342716209676429509660101179587761913570951794712775006017595393099131542462929920832865544705879355440749903797967940767833598657143883346150948256232023103001435628434505839331854097791025034667912357133996133877280328143c = 2044619806634581710230401748541393297937319e = 3<span>m = gmpy.root(c, 3)[0]if pow(m, 3, n) == c:    print long_to_bytes(m)</span>--------------------------------------------------------Output: so_low