PLEASE HELP!!! 14 MORE PROBLEMS TO DO!!!!

Find all residues a such that a is its own inverse modulo 7549.

Question

PLEASE HELP!!! 14 MORE PROBLEMS TO DO!!!!

Find all residues a such that a is its own inverse modulo 7549.

anonymous · Answer

@timo86m ? Did you find anything?

anonymous · Answer

only 1 and -1

anonymous · Answer

$$x^2\equiv 1(\text{mod } p)$$ only if $x=1$ or $-1$ and it turns out 7549 is prime

anonymous · Answer

(Your answer should be a list of integers greater than 0 and less than 7549, separated by commas.)

That's what the problem thing says.

anonymous · Answer

ok

anonymous · Answer

so I guess the only number is 1?

anonymous · Answer

no also i think -1 which means 7548

anonymous · Answer

it says integers greater than 0. but it also says a list so...

anonymous · Answer

AUGH!!

anonymous · Answer

do you know what the question is asking?

anonymous · Answer

yes.

anonymous · Answer

I just don't know how to do it. *sigh*

anonymous · Answer

so  then the list is all integers $n$ less than 7549 with $n^2\equiv 1$ mod 7549

anonymous · Answer

it is a fact that if $p$ is a prime, which 7549 is, that the only solutions are 1 and -1

anonymous · Answer

-1 in this case is the same as 7548

anonymous · Answer

right!! I see what you're getting at now!

anonymous · Answer

TYSM!!!

anonymous · Answer

the proof is more or less straight forward, but you don't need a proof, just the two numbers right?

anonymous · Answer

yw

anonymous · Answer

right. :)