-1

We wish to show that $5^{e}+6^{e} \equiv 0 ($mod$11)$ for all odd numbers $e$, but not for any even numbers $e$.

This comes from Childs' book "A Concrete Introduction to Higher Algebra."

I'm not too sure where to even start, so I would love to make a little headway and show some effort, but I'm genuinely stumped on this one. Haven't done hardly any proofs involving modular arithmetic. Can someone help me out, or at least provide a bit of a startup?

Alessio K
  • 10,599

1 Answers1

2

$$5^e+6^e\equiv 5^e+(-5)^e\equiv 5^e(1^e+(-1)^e) \bmod 11$$

TonyK
  • 64,559
  • Thank you so much! This helps. –  Sep 24 '20 at 11:25
  • Please strive not to post further dupe answers to dupes of FAQs.For site organization is is better to instead link to prior answers (which hopefully have been polished over time due to feedback). Further it helps to localize information on a topic. – Bill Dubuque Sep 24 '20 at 11:35
  • @Bill, the duplicate question that you link to was closed six years ago as a duplicate to another question, which was closed five years ago as lacking focus. Please strive to be more careful what you link to. – TonyK Sep 24 '20 at 12:05
  • There are probably hundreds of (abstract) dupes, and - by design - I chose one that leads to many (cf. "Linked" questions lists) All the more reason not to add further such Q&As with no novelty. Keep in mind that it is typically much more work (and less rewarding) to find (good) dupe targets than it is to give quick short answers like above. Of course you can also add links too. – Bill Dubuque Sep 24 '20 at 19:31
  • 1
    @Bill, please strive not to dupe your very own comments like this. It fills up the site with superfluous verbiage. – TonyK Sep 24 '20 at 20:24
  • @TonyK Comments that attempt to persuade others to help improve the health of the site are by no means "superfluous verbiage". But not to worry, there will be no more here since, alas, they seem to be not well-received. – Bill Dubuque Sep 24 '20 at 20:36