Thread: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

1. Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

Suppose $\displaystyle m$ and $\displaystyle n$ are relatively prime positive integers; show that
$\displaystyle m^{\phi(n)} + n^{\phi(m)} \equiv 1 \pmod{mn}$
where $\displaystyle \phi$ is the Euler Totient function.

I can only see that $\displaystyle \phi(mn) = \phi(m) \phi(n)$ because $\displaystyle gcd(m,n) = 1$.

2. Re: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

Originally Posted by math2011
Suppose $\displaystyle m$ and $\displaystyle n$ are relatively prime positive integers; show that
$\displaystyle m^{\phi(n)} + n^{\phi(m)} \equiv 1 \pmod{mn}$
where $\displaystyle \phi$ is the Euler Totient function.

I can only see that $\displaystyle \phi(mn) = \phi(m) \phi(n)$ because $\displaystyle gcd(m,n) = 1$.
Since $\displaystyle (m,n)=1$ ,we can find k and l such that $\displaystyle km+ln=1$ (Using Extended Euclid's Theorem)
Also using Euler's Theorem,$\displaystyle m^{\phi(n)} \equiv 1 \pmod{n}$ and $\displaystyle n^{\phi(m)} \equiv 1 \pmod{m}$
Can you use the above two facts and take it from there?If not,I will further elaborate.

3. Re: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

Are you aware of the Chinese remainder theorem?
It fits kind of neatly, since it says there's a natural isomorphism between (mod mn) and the combination of (mod m) and (mod n).

4. Re: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

Thank you for your replies. However, I cannot figure out the rest of the solution from either of your hints. Could you please elaborate?

5. Re: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

The Chinese remainder theorem says that the function
f: Z/mnZ $\displaystyle \to$ Z/mZ x Z/nZ
given by
(x mod mn) $\displaystyle \mapsto$ (x mod m, x mod n)
is an isomorphism if m and n are relatively prime.

Please give some feedback on what you know or do not know, or perhaps what you've tried.

Pick $\displaystyle x=m^{\phi(n)} + n^{\phi(m)}$.

6. Re: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

I can solve it now. Thank you for your help.

$\displaystyle x=m^{\phi(n)} + n^{\phi(m)}$

From $\displaystyle x \equiv 1 \pmod{m}$ we have
$\displaystyle x = 1 + my$.
Substitute into $\displaystyle x \equiv 1 \pmod{n}$ we get
$\displaystyle 1 + my \equiv 1 \pmod{n}$.
which simplies to
$\displaystyle my \equiv 0 \pmod{n}$.
Solve $\displaystyle y$ to get
$\displaystyle y \equiv 0 \pmod{n}$,
$\displaystyle y = nz$.
Substitute back into first equation to get
$\displaystyle x = 1 + mnz$.
Hence:
$\displaystyle x \equiv 1 \pmod{mn}$.

7. Re: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

Hmm, I can't follow what you did there (without making a great effort).

I'll just show my proof (using the definition of f from my previous post):
$\displaystyle x \mod m \equiv m^{\phi(n)} + n^{\phi(m)} \mod m \equiv 1 \mod m$
$\displaystyle x \mod n \equiv m^{\phi(n)} + n^{\phi(m)} \mod n \equiv 1 \mod n$

Therefore:
$\displaystyle f: (x \mod {mn}) \mapsto (1 \mod m, 1 \mod n)$

Since we also have from the definition of f:
$\displaystyle f: (1 \mod {mn}) \mapsto (1 \mod m, 1 \mod n)$

and since f is bijective (it's an isomorphism), it follows that
$\displaystyle x \mod {mn} \equiv 1 \mod {mn}$

$\displaystyle \square$

8. Re: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

Did you mean that the definition of $\displaystyle f$ is $\displaystyle f : (x \pmod{mn}) \to (x \pmod{m}, x \pmod{n})$? Why is this a bijective function? Do we need to prove that it is a bijective function?

P.S. The method I used is one of the methods for solving simultaneous congruences.

9. Re: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

Originally Posted by math2011
Did you mean that the definition of $\displaystyle f$ is $\displaystyle f : (x \pmod{mn}) \to (x \pmod{m}, x \pmod{n})$? Why is this a bijective function? Do we need to prove that it is a bijective function?

P.S. The method I used is one of the methods for solving simultaneous congruences.
Yes it is.
The function f is bijective because the Chinese remainder theorem says so.

Do we need to prove it?
I don't know.
You have not set any boundaries on what you can or cannot use.

The Chinese remainder theorem also says exactly how you can solve a simultaneous set of congruences.
See for instance this link for a simple formula.

10. Re: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

Thank you for helping me with this problem. I really appreciate it.

11. Re: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

You're welcome.

12. Re: Suppose gcd(m,n)=1, show that m^{phi(n)} + n^{phi(m)} = 1 (mod mn)

Hello,

Another way : modular arithmetic

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if m and n are relatively prime proved that m^phi(n) n^phi(m)=1

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