Jason Hanson Jokes About Tom Brady Possibly Matching His NFL Milestone

Former Detroit Lions placekicker Jason Hanson joked that he will "fade into obscurity" if Tom Brady matches his record for the most seasons with the same franchise in the NFL.

Brady has spent 20 campaigns with the New England Patriots, and while there is speculation regarding his future ahead of potential free agency, it would be a shock if he didn't extend his stay at Foxborough into a 21st season.

Speaking to Mike Reiss of ESPN, Hanson spoke about Brady nearing the milestone he reached in Detroit:

"I did know it. Nobody cares that I have the record, but now that Tom is close—and ESPN puts it up there—I have friends who take screenshots and they send it to me. And they remind me that I'm about to fade into obscurity. So I'm taking all the accolades for it while it lasts.

"...Obviously, if and when he does that, he will be the focus of that record. And he could go beyond it."

Brady's previous year with the Patriots put him level with Darrell Green and Jackie Slater, who spent two decades with the Redskins and Rams, respectively.

Gregory Shamus/Getty Images

The 42-year-old has yet to secure another season in New England, though, and from Monday at 12 p.m. ET, other teams will be able to negotiate freely with Brady.

It's been reported the Tampa Bay Buccaneers are ready to make a move:

"

The Buccaneers are going “all in” to sign Tom Brady, with head coach Bruce Arians leading the way, per @NFLSTROUD.

Tampa Bay has $80M in cap space  pic.twitter.com/1cwX1ORcXI

— B/R Gridiron (@brgridiron) March 11, 2020"

According to WEEI's Dale Arnold, the Patriots want to bring Brady back next season, although they would not be willing to pay him as much as he made in 2019 ($23 million).

While he is considered by many to be the greatest quarterback of all time, last season was a tough one for Brady and the Patriots by their own high standards. Their playoff chances were dashed early, as they lost at home to the Tennessee Titans in the Wild Card Round.