Simple but annoying characteristic of congruences

[trap101]

I've been having this small problem with modulo arithmetic that's just irking me and this is the perfect example to get clarifivation

find the remainder for 7⁵ congruent to x (mod 21)


here's my issue...so I'd set this up to solve and see if I could find this relationship

7\equiv x (mod 21)...now 7 is a multiple of 21, but of course 21 doesn't divide 7 unless we take into account the negative numbers which would produce 7 \equiv -12 (mod 21)...is this the right way to look at that?

the next thing I did and which made it easier was look at 7² \equiv 7 (mod 21) ...now I have 7⁵ \equiv 7 (mod 21) when all is said and done...but the fact that I know 7 has a relationship with 21 is sitting uneasy with me...

 

[Dick]

Don't know what's sitting uneasy with you. Sure, if 7^2=7 (mod 21) then 7^5=7 (mod 21).

 

[trap101]

Don't know what's sitting uneasy with you. Sure, if 7^2=7 (mod 21) then 7^5=7 (mod 21).

Well it has to do with an RSA encoding question, in order to encode my message which in this case is 7, I have to send the remainder when I raise 7 to the power of 5...which ended up being 7. Probably I'm just over thinking it