I am a not a math major. Have tried with some of the first $1000$ primes. Has not failed.
Asked
Active
Viewed 47 times
0
-
All numbers that are not multiples of 3, when squared, give one more than a multiple of 3 as a result of the squaring. When you subtract 10 from that ... . – Oscar Lanzi Jul 27 '20 at 19:54
-
1Every prime greater than $3$ is of the form $3n+1$ or $3n+2$. (Why?) Now square these expressions, subtract $10$, and you will find that both contain a factor of $3$. – TonyK Jul 27 '20 at 19:56
-
Does this answer your question? The square of a prime is one greater than a multiple of 24? – KingLogic Jul 27 '20 at 19:58
-
1$P$ does not have to be prime for $3$ to divide $P^2-10$; $P$ just cannot be a multiple of $3$ – J. W. Tanner Jul 27 '20 at 20:02
4 Answers
2
Such primes aren't divisible by $3$. But$$3|p\pm 1\implies 3|(p+1)(p-1)=p^2-1\implies3|p^2-1-3\times3=p^2-10.$$
J.G.
- 115,835
2
$p=3n+1$ or $p=3n+2$,
so $p^2=9n^2+6n+1=3(3n^2+2n)+1$ or $p^2=9n^2+12n+4=3(3n^2+4n+1)+1$;
in any event, $p^2=3m+1$, so $p^2-10=3m-9=3(m-3)$ is a multiple of $3$.
J. W. Tanner
- 60,406
2
First, if $P^2-10$ is divisible by 3 then so is $P^2-1$ simplifying the problem.
Because prime numbers greater then three don’t divide three they have two options:
- $P$ is divides 3 with reminder 1. then,$P=3k+1$ For some $k$.
If we square it we get $P^2=9k^2+6k+1$ Because $\overbrace{9k^2 + 6k + 1}^{P^2} - 1 = 3(3k^2 + 2k)$ the condition holds. - $P$ is divides 3 with reminder 2. then,$P=3k+2$ For some $k$.
If we square it we get $P^2=9k^2+12k+4$ Because $\overbrace{9k^2 + 12k + 4}^{P^2} - 1 = 3(3k^2 + 4k+1)$ the condition holds.
razivo
- 2,205
1
All primes $>3$ are numbers of the form $p=6k\pm 1$. so $p^2=36k^2 \pm 12k +1$ and $p^2-10=36k^2 \pm 12k -9 =3(12k^2 \pm 4k -3)$, i.e. a multiple of $3$.
Keith Backman
- 7,288