The product of two sequential primes, minus 1, is prime.
Is this true?
The product of almost every pair of primes is going to be odd, so subtracting one will make it even (and not prime).
3 and 5 are sequential primes.
3 * 5 = 15
15 - 1 = 14
14 is not prime.
So no, it isn’t true.
Brian
Yes. The proposition is only true for the first two primes, 2 and 3, where you get 5.
You forgot to subtract 1. The result is 4 and hence, not prime. I wonder if the OP phrased the question incorrectly because, as written, it seems to fail every time.
2*3=6
6-1=5
Thank you! I do think I’m reasonably good at easy mental arithmetic, like 2 times 3!
Yes, I’m wondering too what proposition the OP is thinking of. Perhaps it’s the proof that there is an infinite number of primes, because if there are only n primes, then:
(P1 * P2 * P3 * … * Pn) - 1
is an integer which won’t have any on the n primes as a factor, and so is itself prime, or has a prime factor distinct from those n primes.
I know there is the (unproven) Goldbach Conjecture which states that there are an infinite number of “prime pairs” (primes seperated by a difference of 2 - e.g. 11 & 13; 29 & 31, etc).
Its “corollary” states that any even number (greater than 6) is the sum of 2 primes. (Perhaps the latter is being referred to by the OP?)
The Goldbach conjecture is that every even number greater than 2 is expressible as the sum of two (not necessarily distinct) primes. The conjecture that there are an infinite number of pairs (p, p + 2) with p and p + 2 prime is known rather unpoetically as the twin prime conjecture.
Thanks ultrafilter.
It’s just that every time the Goldbach Conjecture is discussed, the twin prime conjecture usually enters into the conversation. I wasn’t sure of a definite connection between the two and that’s why I put “corollary” in quotes.
And what happened to Yeppa who started this thread?
Well, this value is prime. It does not have other prime factors, whether distinct from the constituent primes or not. If it had any such factor, it would be composite, and the proof wouldn’t work.
Anyway, I too share in the confusion of what the OP was referring to.
Not the “product”. I thought it was the “sum” of any two sequential primes, etc etc (excluding the first set, 2 & 3).
No – the first counter-example is 17 + 19 - 1 = 35 = 5 x 7; the second is 23 + 29 - 1 = 51 = 3 x 17
No: the product of the first four primes, minus 1 (2357-1) is 209, and that is composite (1123).
209 = 11 * 19. But noted.
This, however, raises a different issue. The famous proof you’re referring to examines the value (P1 * P2 * … * Pn) + 1 — plus, not minus. The whole punch-line of this proof is that whatever finite set of primes you might have, it cannot possibly be covering all of them, because you can always generate a new prime from the members of the set. Therefore, the number of all primes is infinite.
Since (P1 * P2 * … * Pn) – 1 sometimes produces composite numbers, it’s unhelpful for reaching this conclusion.
11 * 13 + 1 = 144, which is most certainly not prime. I think you’ve misunderstood the proof.
Almost indeed. . . .Every prime number except 2 is odd, and the product of two odd numbers is odd, so the product of any two non-2 prime numbers is odd.
Euclid’s proof just has to show that there exists some prime not on your finite list; it does not have to specify what it is. The climax of the proof is “Either N is prime, or N has a prime factor that is not on the original list. Either way, there exists another prime”. For this purpose, adding or subtracting one would both be equally effective. Euclid’s version of the proof happened to use plus, but that’s irrelevant.
I think that having +1 rather than -1 makes it a bit easier to be sure that you have an integer greater than 1 and distinct from the other primes, without needlessly complicating the proof. But if you assume that P1 = 2 and P2 = 3, then N must be not less than 5, so -1 will work except in the most trivial of cases (i.e., where you only know P1 = 2, and the product of known primes - 1 = 1).
Mea culpa, and apologies to Giles. I’m guilty of the mistake described on this page, after the first “Proof”.