I have this number:

84586075770624998147441974158189380700174460760744963832062282657

I generated it using this:

16515489654164894**4 + 10046498489496563**4

so its not square root (power of 2). Its power of 4.

Question: Its possible to compute which power of 4 form my original number and get back orig number? Its possible to do that using power of 2, but what about power of 4?

MathKing Apr 14, 2020

#1**+1 **

sqrt(84586075770624998147441974158189380700174460760744963832062282657, 4 = 17053939551218935.1675822943023889

do you mean this? this is the 4th root of your gigantic number using web 2.0 calculator and the \(\sqrt[y]{x}\) button...... put in 'x' push the button then enter 'y' ans =

ElectricPavlov Apr 14, 2020

#2**0 **

If you expect the fourth root of your large number to be an integer you are doomed to disappointment - search "Fermat's last theorem".

Alan Apr 14, 2020

#3**0 **

Not exactly. Its possible to compute square of 2 (269916694664582966990754893514409**2 + 108309988975954046249057727301124**2) , but why it´s not possible to compute higher (such as 4)? It works for 2, so it must also work with 4.

MathKing Apr 14, 2020

#6**0 **

MathKing, you are not only doomed to disappointment, you are just doomed.

**Not all is lost:** If you and Mr. BB have one of your “intellectual” conversations, then maybe Ginger will write some more hilarious satire.

Like this: https://web2.0calc.com/questions/who-noticed-this-post-by-admin#r2 LMAO!

Guest Apr 14, 2020

#9**0 **

The expectation for finding an exception to Fermat’s Last Theorem is zero. The proof for that is almost 30 years old. Ginger comments on it here: https://web2.0calc.com/questions/help_79263#r8

As for the expectation that Ginger will write more satire, it probably depends on how your conversation with Mr. BB unfolds. The one you had a year ago was new and obviously inspiring. However, Ginger is very talented. She often can make silk purses from sow’s ears: it’s worth a shot

Guest Apr 14, 2020