Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss

Find all primes of the form 4^n + n^4

40 views
Skip to first unread message

david

unread,
Nov 10, 2003, 11:52:06 AM11/10/03
to
A problem:
* Find all primes of the form 4^n + n^¤

I just can't do it!

4^1 + 1^4 = 5, is prime
* If 2 divides n, 2 divides 4^n + n^4 so it can't be prime.

* If 2 does noes not divide n and 5 doesn't either 5 divides 4^n + n^4
so it isn't prime. (4^n always ends in 4 for odd n and n^4 always ends
in 1 for odd n that aren't divisible by 5. Thus 4^n + n^4 always ends in
5 and is divisible by 5. If anyone has a more elegant way of proving
this, please let me know.)

* This leaves us odd n that are multipiles of 5. I suspect these are all
compsite too. 4^5 + 5^4 = 1649 = 17*97, at least. But how do you prove it?

/david

Robin Chapman

unread,
Nov 10, 2003, 11:46:27 AM11/10/03
to
david wrote:

I set this as a "challenge problem" in my number theory course:
http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html .

--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Needless to say, I had the last laugh."
Alan Partridge, _Bouncing Back_ (14 times)

david

unread,
Nov 10, 2003, 1:01:01 PM11/10/03
to

>
> I set this as a "challenge problem" in my number theory course:
> http://www.maths.ex.ac.uk/~rjc/courses/nt03/nt03.html .
>

But you didn't provide a proof in the answer sheet! =(

Robin Chapman

unread,
Nov 10, 2003, 12:55:45 PM11/10/03
to
david wrote:

Heh, heh, heh!

David Einstein

unread,
Nov 10, 2003, 1:17:02 PM11/10/03
to

A short hint. Complete something.

david

unread,
Nov 10, 2003, 2:40:15 PM11/10/03
to
The proof?

David Einstein

unread,
Nov 10, 2003, 3:00:12 PM11/10/03
to
david wrote:

Something else first.

Ignacio Larrosa Cañestro

unread,
Nov 10, 2003, 3:54:18 PM11/10/03
to

"david" <s...@asf.com> escribió en el mensaje
news:FePrb.2430$%W3.18479@amstwist00...

Fon even n is obvious. For odd n, replace n = 2k+1.


--
Best regards,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUIT...@mundo-r.com


Julien Santini

unread,
Nov 10, 2003, 4:56:59 PM11/10/03
to
>
> Fon even n is obvious. For odd n, replace n = 2k+1.
>

For odd n such that 5 does not divide n, this is obvious too ... but what
about the case n = 5*i, where i is odd?


David Einstein

unread,
Nov 10, 2003, 5:05:56 PM11/10/03
to
Forget the divisibility by 5, treat all the odd n the same. (you might
benefit from writing one of the 4s differently)

Ignacio Larrosa Cañestro

unread,
Nov 10, 2003, 5:11:20 PM11/10/03
to

"Julien Santini" <santini...@wanadoo.fr> escribió en el mensaje
news:bop1jc$pgn$1...@news-reader5.wanadoo.fr...

> >
> > Fon even n is obvious. For odd n, replace n = 2k+1.
> >
>
> For odd n such that 5 does not divide n, this is obvious too ... but what
> about the case n = 5*i, where i is odd?

For n = 2k + 1,

n^4 + 4^n = (n^2 + 2^(k + 1)n + 2^n)(n^2 - 2^(k + 1)n + 2^n)

Julien Santini

unread,
Nov 10, 2003, 5:18:02 PM11/10/03
to
> For n = 2k + 1,
>
> n^4 + 4^n = (n^2 + 2^(k + 1)n + 2^n)(n^2 - 2^(k + 1)n + 2^n)
>

Magical !!

Regards,
Julien Santini


David Petry

unread,
Nov 10, 2003, 6:54:59 PM11/10/03
to
david <s...@asf.com> wrote

> A problem:
> * Find all primes of the form 4^n + n^¤

Borrowing one of Einstein's ideas, we have
4^N+N^4 = (2^N+N^2)^2 - 2^(N+1)N^2,
and the rest is relatively easy.

Hugo Pfoertner

unread,
Nov 10, 2003, 7:24:56 PM11/10/03
to

I searched for n^4+4^n = semiprime, i.e.
(n^2 + 2^(k + 1)n + 2^n) and (n^2 - 2^(k + 1)n + 2^n) both prime
and found only n=3,5,15,35,55
e.g. 3^4+4^3=145=5*29, 5^4+4^5=1649=17*97, ...

can we find more terms or is there a fundamental reason that no
semiprime representations exist beyond n=55?

Hugo Pfoertner

Dave Rusin

unread,
Nov 10, 2003, 8:23:06 PM11/10/03
to
In article <FePrb.2430$%W3.18479@amstwist00>, david <s...@asf.com> wrote:

>* Find all primes of the form 4^n + n^¤

[This line ends with a non-ASCII character after a "^"]

>* If 2 divides n...
>* If 2 does noes not divide n...

This is one of those funny cases where a problem gets easier when
you make it harder: I claim at that point you're actually looking
at the broader question, "Find all primes of the form n^4 + 4 m^4."

Just for the record, I seem to recall that the polynomial n^4 + k^2
was recently proven to yield infinitely many prime values (as opposed
to the polynomial j^2 + k^2, which is easy, and the polynomial
1 + k^2, which is still unknown).

dave

Robert Israel

unread,
Nov 10, 2003, 10:19:28 PM11/10/03
to
In article <3FB02C58...@abouthugo.de>,
Hugo Pfoertner <not...@abouthugo.de> wrote:

>I searched for n^4+4^n = semiprime, i.e.
>(n^2 + 2^(k + 1)n + 2^n) and (n^2 - 2^(k + 1)n + 2^n) both prime
>and found only n=3,5,15,35,55
>e.g. 3^4+4^3=145=5*29, 5^4+4^5=1649=17*97, ...

>can we find more terms or is there a fundamental reason that no
>semiprime representations exist beyond n=55?

It's quite reasonable that there should be only finitely many.
Heuristically the probability of a number the size of these being
prime is on the order of 1/n. The probability of both being prime
would be on the order of 1/n^2. Since sum_n 1/n^2 < infinity, we
should expect only a finite number of examples.

Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2

Hugo Pfoertner

unread,
Nov 11, 2003, 12:01:52 AM11/11/03
to

There are no further solutions up to k=5000 -> n=5001. So the practical
chances of finding another solution (if one exists) are rather limited.

Hugo Pfoertner

Hugo Pfoertner

unread,
Nov 11, 2003, 12:03:52 AM11/11/03
to
Hugo Pfoertner wrote:
>
> There are no further solutions up to k=5000 -> n=5001. So the practical
n=10001 sorry
0 new messages