3

$k^2+1$ has a prime divisor bigger that $2k$.

For instance, if $k=5$ we have $k^2+1=26$ where $13$ is a prime divisor bigger than $2k=10$.

I try to follow Euclid's proof. By absurd, suppose that the set of the positive integers such that $k^2+1$ has a prime divisor bigger than $2k$ is finite.

We denote : $k_1,...,k_n$ these numbers.

Now we want to build $k_{n+1}$ which depends on the others and checks the property. But I don't know how to continue.

Thanks in advance !

Maman
  • 3,478

1 Answers1

2

Suppose, for the sake of contradiction, that only $k_1,\ldots,k_n$ satisfy the requirements. Now let $q_1,\ldots,q_m$ be all prime numbers dividing at least one of the numbers $k_1^2+1,\ldots,k_n^2+1$ and define $N:=(2q_1\cdots q_m)^2+1$ and let $p$ be any prime factor of $N.$ Suppose that $2q_1\cdots q_m\equiv r\pmod p,$ with $1\leqslant r\leqslant p-1$ and set $M:=\min\{r,p-r\}.$ Note that $p-r\geqslant p/2$ if and only if $p/2\geqslant r.$ Then $M^2+1\equiv0\pmod p$ and also $M\leqslant p/2$ however, since $N$ is odd, then $p$ is odd so $M<p/2$ or equivalently $2M<p.$ It follows that $M=k_i$ for some $i,$ which implies that $p=q_j$ for some $j,$ which is clearly false.

CIJ
  • 3,477
  • 1
    You don't really need the part which uses $q_1,\ldots, q_m$. You just need to argue that $k^2+1$ can have arbitrarily large prime factors, so that $p$ can be taken arbitrarily large. So $N = (n!)^2+1$ would suffice (alternatively, use the fact that the set of $r$-smooth numbers is very sparse for any fixed $r$, much sparser than the sequence $k^2+1$). – Erick Wong Nov 22 '16 at 20:32
  • Why did you introduce the set $M$ ? – Maman Nov 22 '16 at 20:37
  • @Maman I introduce the number $M$ because I wanted to find some positive integer $l$ such that $l^2+1$ had a prime factor $p>2l$ and $l\neq k_i$ for all $i.$ At that moment, it was the first thing that came to my mind given the conditions of the problem and the fact that I couldn't see a way to prove that $N$ has a prime factor $p>4q_1\cdots q_m.$ – CIJ Nov 22 '16 at 20:48
  • Do we have $M^2+1=N$ ? In that case that's why $M^2+1\equiv 0 \pmod p$ ? – Maman Nov 24 '16 at 22:33
  • No... By definition, $M$ is the minimum of $r$ and $p-r,$ where $r$ the residue of $2q_1\cdots q_m$ when this number is divided by $p$ (where $p$ is any prime factor of $N$). As I mentioned in my answer, $M$ must satisfy $2M<p.$ Now, if $M=r$ then $M^2+1\equiv r^2+1\equiv(2q_1\cdots q_m)^2+1=N\equiv0\pmod p$ because by definition $p\mid N.$ Something similar happens if $M=p-r.$ – CIJ Nov 24 '16 at 22:59