0

For any odd integer $a$, show that $a^{33} \equiv a \pmod{4080}$. I have attempted this problem in the following way:

Observe that $4080 = 2^4 \times 3 \times 5 \times 17 = 2 \times 15 \times 4 \times 2 \times 17$.\ $\bullet$ Since $a$ is odd, so $\gcd(a, 2\times 15) = 1$. Thus, $a^{\phi(30)} \equiv 1 \pmod{30}$ along with $$\phi(30) = \phi(2 \times 3 \times 5) = 2 \times 3 \times 5 \times (1 - \frac{1}{2})(1 - \frac{1}{3}) (1 - \frac{1}{5}) = 30(\frac{1}{2})(\frac{2}{3}) (\frac{4}{5}) = 8$$ yields $a^8 \equiv 1 \pmod{30}$ i.e. $a^{32} \equiv 1 \pmod{30}$.\ $\bullet$ Since $a$ is odd, so $\gcd(a, 2\times 17) = 1$. Thus, $a^{\phi(34)} \equiv 1 \pmod{34}$ along with $$\phi(34) = \phi(2 \times 17) = 2 \times 17 \times (1 - \frac{1}{2})(1 - \frac{1}{17}) = 34(\frac{1}{2})(\frac{16}{17}) = 16$$ yields $a^{16} \equiv 1 \pmod{34}$ i.e. $a^{32} \equiv 1 \pmod{34}$.\ $\bullet$ Since $a$ is odd, so $\gcd(a, 16) = 1$. Thus, $a^{\phi(16)} \equiv 1 \pmod{16}$ along with $$\phi(16) = \phi(2^4) = 2^4 - 2^3 = 8$$ yields $a^8 \equiv 1 \pmod{16}$ i.e. $a^{32} \equiv 1 \pmod{16}$.\ Now, $\rm{lcm}(30, 34, 16) = 4080$, so from these congruences, we have $$a^{32} \equiv 1 \pmod{\rm{lcm}(30, 34, 16)} \equiv 1 \pmod{4080}$$ and hence $a^{33} \equiv a \pmod{4080}$. \end{enumerate}

The issue is that in "Since $a$ is odd, so $\gcd(a, 2\times 15) = 1$", if we choose $a = 3$, an odd integer, then $\gcd(a, 2\times 15) \neq 1$.

Bill Dubuque
  • 272,048
Khalid Naeem
  • 73

1 Answers1

2

Hint $\ 3\mid a \,\Rightarrow\, 3\mid a^{\large 33}\!-a.\,$ Ditto for $\,p=5,17.\,$ So we can reduce to the case that $\,a\,$ is coprime to $2,3,5,17,\,$ i.e. $\,a\,$ is coprime to $4080,\,$ where we can combine the Euler results at each prime.

Remark $\ $ See also Carmichael's Lambda function, a generalization of Euler's phi function.

Bill Dubuque
  • 272,048