Evenly divisible

Where % ~ mod, it looks like you are using these principles:
More generally, you could write:
(n+1)^(n-1)  % n = [ (n+1) % n ] ^ (n-1) = [ n % n +1 % n ] ^ (n-1) = [ 0 + 1 ] ^ (n-1) = 1
(n-1)^(n+1)  % n = [ (n-1) % n ] ^ (n+1) = [ n % n - 1 % n ] ^ (n+1) = [ 0 - 1 ] ^ (n+1) = -1

So to get the correct answer. Would I use this
(123^121+121^123) mod 122= ((122+1)^121) mod 122+ ((122-1)^123)mod 122= (1^121) mod 122 +((-1)^123) mod 122 = 1 mod 122+ (-1) mod 122= (1-1) mod 122 =0

or this?
(n+1)^(n-1)  % n = [ (n+1) % n ] ^ (n-1) = [ n % n +1 % n ] ^ (n-1) = [ 0 + 1 ] ^ (n-1) = 1
(n-1)^(n+1)  % n = [ (n-1) % n ] ^ (n+1) = [ n % n - 1 % n ] ^ (n+1) = [ 0 - 1 ] ^ (n+1) = -1

I prefer symbols, but you can choose the approach you feel more comfortable with.

Let n = 122, then
(n+1)^(n-1)  % n = (122+1)^(122-1)  % 122 = (123^121) mod 122

and
(n-1)^(n+1)  % n = (122-1)^(122+1)  % 122 = (121^123) mod 122

Can you make the final conclusions from this substitution by comparing it with your OP?

no :-(

>> no :-(
Could you elaborate?
For example, what part of the substitution are you having trouble with?

well all I really wanted to know was if my solution was correct

Yes.

These are your steps with the justifications. Looks complete to me.

Oh, that first step also used (a+b) % n = a%n + b%n