[david_allison]

> Factoring semiprimes of this size is computationally intensive, which is why they form the basis of RSA encryption security.

Not if it's divisible by 2.

    from sympy import isprime
    num = 29927402397991286489627837734179186385188296382227646249397073654051914085318503794952624411151858464246403027505634195232053330357484129331920822220662818816547063469215394303721576869467659309978113411955550111870966028627418736664
    print(num//2) # 14963701198995643244813918867089593192594148191113823124698536827025957042659251897476312205575929232123201513752817097616026665178742064665960411110331409408273531734607697151860788434733829654989056705977775055935483014313709368332
    print(isprime(num//2)) # False

[rowanG077]

Indeed that works for that case. But you can prompt yourselves, it will not always generate natural that are easy to validate with such shortcuts. So I don't think it invalidates the point I'm making.