$\{n\in\mathbb N|n\in\mathbb P \wedge n ^2+2\in\mathbb P\}$ is finite

Question

In an article on Wikipedia there is a claim of a proof that it don't exist infinitely many $n\in\mathbb N$ such that both $n$ and $n^2+2$ are primes. I don't understand that and would be pleased if someone could explain.

Okay, thank you, but it is the text in that section that I don't understand. This is supposed to have something to do with integer-valued polynomials and fixed prime divisors.

Answer 1

Now I finally understand what Wikipedia was trying to bring about. The subring of all integer-valued polynomials with rational coefficients is a free Abelian group where all elements $f$ can be uniquely expressed as linear combinations of the (in this ring) irreducible polynomials

$\displaystyle \left( \begin{matrix} X \\ k \\ \end{matrix}\right)= \frac{X(X-1)(X-2)\cdots(X-k+1)}{k!}$
where $c_0=f(0)$ and $\displaystyle c_k=f(k)-\sum_{i=0}^{k-1}c_i {k\choose i}$.

Now $|\gcd(c_0,\dots,c_n)|$ is the greatest number that divides all outputs $f(x)$ for $x\in\mathbb Z$.
In the example $x(x^2+2)$: $\;c_0=0,\;c_1=3,\;c_2=6,\;c_3=6$.

Answer 2

Consider some prime integer $n \geq 4$. Then we can write $n=3k+r$, where $k$ and $r$ are positive integers ($3$ cannot divide $n$), and $r$ is $1$ or $2$.

Then $n^2+2=(3k+r)^2+2=3(3k^2+2rk)+(r^2+2)$.

You can check that $r^2+2$ is always divisible by $3$; therefore $n^2+2 \geq 18$ is divisible by $3$, thus cannot be prime.

Answer 3

If $n$ and $n^2+2$ are both prime, then since $3|n(n^2+2)$, either $3|n$ or $3|n^2+2$. If also $n$ and $n^2+2$ are prime, then either $n = 3$ or $n^2 + 2 = 3$ (so $n=1$, which is not prime). Thus, there is only one (positive) integer prime $n$ such that $n^2+2$ is prime: it is $3$.

Answer 4

Read the page better: $(n-1)n(n+1)$ is divisible by $3$ as it is a product of three consecutive integers and also $3n$ is divisible by $3$, by definition.

And so $3$ also divides their sum:

$$n(n+1)(n-1) + 3n = n(n^2-1) + 3n=n^3 -n + 3n = n^3 + 2n = n(n^2+2)$$

and so $3$ either divides $n$ or it divides $n^2+2$. So it cannot be that both are primes, except when $n=3$ itself (and we have $3$ and $11$).

Answer 5

$\{n\in\mathbb N\mid n\in\Bbb P\land n^2+2\in\Bbb P\}=\{3\}$.

This is because $n(n^2+2)=(n-1)n(n+1)+3n$, and the RHS is divisible by $3$. Then by Euclid's lemma, $3\mid n\lor3\mid(n^2+2)$.

Answer 6

All primes other than $2,3$ have the form $p=6k\pm1$. Accordingly, the squares of those primes have the form $p^2=6j+1$, and $p^2+2$ will have the form $6j+3$ which is divisible by $3$ on its face (see comment by lulu to original question). This leaves $2,3$ as the only possible candidates; $2^2+2=6$ and $3^2+2=11$, making $3$ the sole prime satisfying the condition.