$k^2+1$ 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 !
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.