Is (2^82589933-243)/19^2 prime?

[enzocreti]

Could you proof that (2^82589933-243)/19^2 is composite?
No factor below 4*10^9

[Batalov]

Originally Posted by enzocreti

Is (2^82589933-243)/19^2 prime?

Bet you $100 against your 1 lira that it isn't.

[Uncwilly]

How did you pull that number out of your ear or rear?

[CRGreathouse]

Originally Posted by Batalov

Bet you $100 against your 1 lira that it isn't.

I assume your bet is that the number is composite?

[ATH]

Originally Posted by enzocreti

Could you proof that (2^82589933-243)/19^2 is composite?
No factor below 4*10^9

The number is 24.8 million digits long it could have factors that are millions of digits long (up to 12.4 million digits), the fact that it has no factor below 4,000,000,000 barely changed the probably that it is prime at all.
You could trial factor it to 4,000,000,000,000,000,000,000 without any factor, and it would still most likely be composite.

[dcheuk]

Originally Posted by enzocreti

Could you proof that (2^82589933-243)/19^2 is composite?
No factor below 4*10^9

Isn't \(2^{82589933}-1\) a Mersenne Prime? Is it true in general that for each Mersenne Prime \(M\), then \(\dfrac{M-242}{19^2}\) is also prime?

I'm just asking out of curiousity maybe this is a stupid question, but how do we see that this number is in Z?

[GP2]

Originally Posted by dcheuk

I'm just asking out of curiousity maybe this is a stupid question, but how do we see that this number is in Z?

2^82589933 mod 361 = 243

[dcheuk]

Originally Posted by GP2

2^82589933 mod 361 = 243

Sorry kinda slow lol.

\[
\begin{align*}
2^{82589933}&\equiv2^{\left\lfloor82589933/\phi(19^2)\right\rfloor}\pmod{19^2}\\
&\equiv 2^{\left\lfloor82589933/(19^2-19)\right\rfloor}\pmod{19^2}\\
&\equiv 2^{11}\pmod{19^2}\\
&\equiv 243\pmod{19^2}
\end{align*}\]

[GP2]

Originally Posted by dcheuk

\[\equiv 243\pmod{19^2}\]

Most programming languages (other than C and C++) have modular exponentiation library functions.

For instance, in Python: pow(2,82589933,19*19) gives 243

[retina]

Originally Posted by dcheuk

Sorry kinda slow lol.

\[
\begin{align*}
2^{82589933}&\equiv2^{\left\lfloor82589933/\phi(19^2)\right\rfloor}\pmod{19^2}\\
&\equiv 2^{\left\lfloor82589933/(19^2-19)\right\rfloor}\pmod{19^2}\\
&\equiv 2^{11}\pmod{19^2}\\
&\equiv 243\pmod{19^2}
\end{align*}\]

Not sure but isn't it mod instead of floor division?

[dcheuk]

Originally Posted by retina

Not sure but isn't it mod instead of floor division?

Yeah you're right my bad.

To correct mine: \(2^{82589933}\equiv 2^{82589933-\left\lfloor82589933/(19^2-19)\right\rfloor\cdot(19^2-19)}\pmod{19^2}\) should do it.