Can Modulo Calculations Demonstrate These Number Theory Properties?

[1+1=1]

hi, it's me again, i only have 3 tiny questions then i am done asking, i hope!

i need to show that if gcd(a,n)=(a-1,n)=1, then 1+a+a^2...+a^\phi^n^-^1\equiv0 mod n

show (m,n)=1 then m^\phi^n+n^\phi^m\equiv 1 mod (mn)

show if m and k are positive integers then \phi(^k)=m^k-1\phi(m)

what i know so far: the second one can use fermat's little theroem correct? if a==0 mod b and b==0 mod a then => ab==0 mod(ab)

the third one is just playing with my brain, i honestly do not know anywhere to start it.

the first question says what a,n are relatively prime, and a-1,n are also relatively prime. so, if any a raised to a power, that a is == to 0, mod n. can anyone give me a "hint"?

thank you! p.s. does my LaTeX look good? feel free to tell me and all.

 

[1+1=1]

can anyone help me with these? i honestly have no idea how to start any of them... i just need a little "boost"

 

[M98Ranger]

Hi there,

To show that 1+a+a^2+...+a^\phi^n^-^1\equiv 0 mod n, we can use the fact that \phi(n) is the number of positive integers less than n that are relatively prime to n. This means that for any number a relatively prime to n, a^\phi(n) \equiv 1 mod n.

Since gcd(a,n)=1 and (a-1,n)=1, we can rewrite the expression as 1+a+a^2+...+a^\phi^n^-^1 = (a^\phi)^n + (a^\phi)^(n-1) + ... + a^\phi + 1.

Now, using the fact that a^\phi \equiv 1 mod n, we can rewrite the expression as 1+1+...+1, which is \phi(n) number of 1's. Since \phi(n) is the number of terms in the expression, we can rewrite the expression as \phi(n) \equiv 0 mod n. This shows that the original expression is congruent to 0 mod n.

For the second question, we can use Fermat's Little Theorem which states that if (m,n)=1, then m^\phi(n) \equiv 1 mod n. This means that m^\phi(n) is congruent to 1 mod n. Similarly, n^\phi(m) is also congruent to 1 mod m.

Therefore, we can rewrite the expression as m^\phi(n) + n^\phi(m) \equiv 1+1 mod (mn), which is equivalent to 2 \equiv 1 mod (mn). This shows that the expression is congruent to 1 mod (mn).

For the third question, we can use the fact that \phi(k) = k-1 for any prime number k. This means that \phi(m^k) = m^k-1.

Hope this helps! And yes, your LaTeX looks great! Keep up the good work.