Composite $2^{398}+1$ in its prime factors.

Question

Now I haven’t come around such problems in their general form, mostly I have proved if they prime or not rather than composing it in its prime factors.

I don’t know if this may lead somewhere however :

$$4\equiv -1\pmod 5$$ $$4^{199}\equiv -1^{199}\pmod 5$$ $$4^{199}+1\equiv 0\pmod 5$$ $$2^{398}+1\equiv 0\pmod 5$$

Showing 1 of its prime factors is essentially 5, but I don’t know how to continue. I’m also interested in the general case or optimisation of such problems in general. Thanks.

What do you mean when you state that one “of its prime factors is essentially $5$”? — José Carlos Santos, Dec 07 '24 at 10:42
$2^{398}+1 = 4 \times (2^{99})^4 + 1^4 = 4b^4+a^4 = (a^2-2ab+2b^2)(a^2+2ab+2b^2)$ is why this number is composite. — Sarvesh Ravichandran Iyer, Dec 07 '24 at 10:43
This is as Mr. Ravichandran suggests the Sophie Germain identity very useful in arithmetic problem like this one in OP. — EDX, Dec 07 '24 at 10:45
And yes, $5$ is dividing $2^{398}+1$, and it is the highest power of $5$ dividing it. So you should not say "essentially 5". — Dietrich Burde, Dec 07 '24 at 10:47
Complete the square as in the prior linked dupe (and comment) — Bill Dubuque, Dec 07 '24 at 18:52

Dietrich Burde · Answer 1 · 2024-12-07T12:23:54.153

2

As explained in the comments, $$ 2^{398}+1=(1-2^{100}+2^{199})(1+2^{100}+2^{199}). $$ This is enough for pari-gp to factorize it: \begin{align*} 1-2^{100}+2^{199}& =797 \\ & \cdot 1008116715344410461444141839610180239223178503751442552629, \end{align*} and

\begin{align*} 1+2^{100}+2^{199} & =5\cdot 76554648784441 \\ &\cdot 2099073106303095025303885460879717918033130293 \end{align*}

edited Dec 07 '24 at 12:23

answered Dec 07 '24 at 11:10

Dietrich Burde

140,055

$~2^{199} \approx 10^{66}.~$ Do you have access to a super computer? – user2661923 Dec 07 '24 at 11:36
2

@user2661923 Meet the Alpertron. – Sarvesh Ravichandran Iyer Dec 07 '24 at 11:42
@user2661923 Mathematica running on a Raspberry Pi 5 factors $(1-2^{100}+2^{199})$ in about $0.001255$ seconds and $(1+2^{100}+2^{199})$ in about $0.381867$ seconds, so apparently these two numbers are not particularly difficult to factor probably because they both contain some relatively small factors. – Steven Clark Dec 07 '24 at 16:02
Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Dec 07 '24 at 18:52

Composite $2^{398}+1$ in its prime factors.

1 Answers1