Is there always a prime $p$ so that the largest prime factor of $p^2+i$ not exceeding $p$ for $-k\leq i \leq k$? Define $f(n)$=the largest prime factor of $n$. For example, $f(28)=7.$

Question: Is it true that for any given integer $k>0$, we can find a prime $p>\sqrt k$ so that 
  $$f(p^2-k)\leq p,f(p^2-k+1)\leq p,\dots,f(p^2+k)\leq p~?$$

For example, if $k=6,$ we can take $p=660683,$ since $\{f(660683^2+i)\mid i=-6,-5,\dots,5,6\}=\{76493,53441,132137,278413,19289,55057,660683,614333,73483,43177,3049,273929,171553\}.$
Thanks in advance!
 A: I don't have a proof, but I do have a heuristic argument that the statement is true, as well as some asymptotic estimates on the size of $p$ as a function of $k$.
First, define an unusual number to be a number $n$ whose largest prime factor is strictly greater than $\sqrt{n}$.  One of the requirements in your question is related to unusual numbers:
Proposition. Let $k\geq 0$ be an integer, let $p \geq \dfrac{k-3}{4}$ be an odd prime, and suppose that the set
$$
\{p^2-k,\;p^2-k+1,\;\ldots,\;p^2+k\}
$$
contains no unusual numbers.  Then no element of this set is divisible by any prime number greater than $p$.
Proof: Let $n$ be an element of the set, and let $q$ be the largest prime number dividing $n$.  Since $n$ is not unusual, we know that
$$
q \;\leq\; \sqrt{n} \;\leq\; \sqrt{p^2+k} \;\leq\; \sqrt{p^2+4p+3} \;<\; \sqrt{p^2+4p+4} \;=\; p+2.
$$
Since $q$ is prime and $q < p+2$, it follows that $q\leq p$.$\quad\square$
Now, this paper by John G. Kemeny gives asymptotic estimates on the frequency of unusual numbers.  Specifically, if $u(n)$ denotes the number of unusual numbers less than or equal to $n$, then Kemeny proves
$$\
\lim_{n\to\infty} \frac{u(n)}{n} \;=\; \log 2.
$$
That is, the probability that a given number is unusual is about $\log 2 \approx 69.31\%$.  (Note: This result had previously been announced by Schroeppel in an unpublished 1972 memo, but Kemeny's paper seems to be the first published proof.)
Thus, roughly speaking, the probability that the set
$$
\{p^2-k,p^2-k+1,\ldots,p^2+k\}
$$
has no unusual numbers is $(1-\log 2)^{2k}$.  This value does not depend on $p$, so eventually there ought to be a prime number $p$ for which the corresponding set is free of unusual numbers.
Of course, such heuristic reasoning should always be taken with a grain of salt.  Similar heuristic arguments can be made for the Twin Prime Conjecture and Goldbach's Conjecture, and these are major open problems.
By the way, the reasoning above gives us some rough estimates on how large $p$ needs to be as a function of $k$.  In particular, given a value of $k$, there ought to be (with $>50\%$ chance) a corresponding value of $p$ somewhere in the first $(1-\log 2)^{-2k}\log 2$ primes.  This predicts that there should  to be a value for $k=7$ in the first $11$ million primes, and a value for $k=8$ in the first $112$ million primes.
Edit: After a lengthy computer search, I have found that $83\text{,}323\text{,}327$ works for $k=7$, and $383\text{,}155\text{,}211$ works for $k=8$.  (Primes for $k=1,\ldots,6$ were given in the comments above.)  These numbers are the $4\text{,}851\text{,}928$'th and $20\text{,}485\text{,}208$'th primes, respectively, which agrees roughly with the asymptotic predictions above, although both arrived a bit early.  The prediction for $k=9$ is that it will be satisfied by one of the first $1.2$ billion primes.
