0
$\begingroup$

Is it true that no prime larger than $241$ can be made by either acting or subtracting $2$ coprime numbers made up out of the prime factors $2,3,$ and $5?$

Update

Above example is clearly wrong, as shown by MJD. New question:Is it true that $251$ cannot be made by either acting or subtracting $2$ coprime numbers made up out of the prime factors $2,3,$ and $5?$

$\endgroup$
  • $\begingroup$ Is $1$ allowed as one of the coprime numbers? $\endgroup$ – Mark Bennet Oct 31 '14 at 14:32
  • $\begingroup$ @MarkBennet No ;) $\endgroup$ – martin Oct 31 '14 at 14:34
  • $\begingroup$ What is the basis for your claim? $\endgroup$ – Ali Caglayan Oct 31 '14 at 14:37
  • $\begingroup$ $251 = 256 -5 = 8 +243 $. $\endgroup$ – MJD Oct 31 '14 at 14:51
  • 1
    $\begingroup$ As far as I can tell that is true. $\endgroup$ – MJD Oct 31 '14 at 15:27
4
$\begingroup$

You mean like $162 + 625 = 787$?

Intuitively, it would very surprising if this conjecture were true. There is no reason to expect that the sum or difference of two arbitrary numbers would not in general be prime, and there are quite a lot of primes, so something surprising would have to happen for this large family of sums and differences to almost completely miss all the primes.

Brute-force computer search finds many counterexamples; for example $2^{19} + 3^4 =524369$. If you are interested in this kind of conjecture, learning a minimal amount of computer programming would be a good investment of your time.

$\endgroup$
  • $\begingroup$ Yes, I posted the question in haste, I am afraid. However, there certainly appears to be a list of primes greater than 241 that cannot be made. $\endgroup$ – martin Oct 31 '14 at 14:49
  • $\begingroup$ Also $1039, 2063, 4111, 32783, 65551$ of the form $2^n+15$ $\endgroup$ – Mark Bennet Oct 31 '14 at 14:50
  • 2
    $\begingroup$ @martin It's not surprising; the set of numbers that are sums or differences of 5-smooth numbers has asymptotic density 0, so one should expect its intersection with the primes to be 'small'. But there's no reason to expect it to be finite. $\endgroup$ – Steven Stadnicki Oct 31 '14 at 15:17
  • 1
    $\begingroup$ (A minor correction to my previous comment: the set of sums of 5-smooth numbers has density 0, but I doubt it's known whether the set of differences of 5-smooth numbers does. But I believe that this would follow from the ABC conjecture, so it's fairly plausible. $\endgroup$ – Steven Stadnicki Oct 31 '14 at 15:27
  • $\begingroup$ @StevenStadnicki thanks for the note - I feel slightly embarrased that I didn't check this properly, and I shall take more time in formulating another question in a similar vain! $\endgroup$ – martin Oct 31 '14 at 15:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.