1) f(x)=38x+2

2) f(x)=33x+2

3) x

^{30}

Ok so i am not trying to get the answer here without sweating for it. Warning though, i am terrible at maths.

I took case 1. f is a bijection if f is both injective and surjective.

I started by proving that f is a injection. For that i prove that f(a)=f(b)

f(a)=f(b)

38a+2=38b+2

38(a-b)=0

a=b if a-b=0 so i'm trying to find 38's inverse in Z/77Z

(-2)*38=-76 so the inverse is (-2)

-2(a-b)=0 => a-b=0 => a=b

So f is an injection

Now i want to prove that f is a surjection

y=38x+2

y-2=38x

I'm stuck here, what should i try now.