We're calling on all EU-based Mozillians with iOS or iPadOS devices to help us monitor Apple’s new browser choice screens. Join the effort to hold Big Tech to account!

সহায়তা খুঁজুন

Avoid support scams. We will never ask you to call or text a phone number or share personal information. Please report suspicious activity using the “Report Abuse” option.

আরও জানুন

Why do I need to have an account to use Ffx for Android mobile? I have never used an account for any Firefox browser.

  • 2 উত্তরসমূহ
  • 1 এই সমস্যাটি আছে
  • 9 দেখুন
  • শেষ জবাব দ্বারা James

more options

Why do I need to have an account to use Ffx for Android mobile? I have never used an account for any Firefox browser.

Seems I can't even ask the question without opening an account! Catch 22 defined.

Why do I need to have an account to use Ffx for Android mobile? I have never used an account for any Firefox browser. Seems I can't even ask the question without opening an account! Catch 22 defined.

সমাধান চয়ন করুন

You don't need a Firefox account to use Firefox, you only need it if you want to use the Sync service to synchronize bookmarks and other data between your desktop installation of Firefox and your mobile installation. (I don't personally use Sync, since the desktop installation has a lot of old baggage that would bog down my mobile.)

প্রেক্ষাপটে এই উত্তরটি পড়ুন। 👍 0

All Replies (2)

more options

চয়ন করা সমাধান

You don't need a Firefox account to use Firefox, you only need it if you want to use the Sync service to synchronize bookmarks and other data between your desktop installation of Firefox and your mobile installation. (I don't personally use Sync, since the desktop installation has a lot of old baggage that would bog down my mobile.)

more options

MorgansDaddi00 said

Seems I can't even ask the question without opening an account! Catch 22 defined.

This forum needs an account to post in yes. They did allow Guest posting for a year or two about but it was hard to know who was whom and well had cases of spam and trolls also.

James দ্বারা পরিমিত