prove that $2^{15} - 2^3 $ divides $ a^{15} - a^3$

Question

Prove that $$2^{15} - 2^3 $$ divides $$ a^{15} - a^3$$ for any integer $a$. Hint: $$ 2^{15} - 2^3 = 5\cdot7\cdot8\cdot9\cdot13$$

Answer 1

Hint: If $13\mid a$, then $13$ divides $a(a^{14}-a^2)=a^{15}-a^3$. Otherwise, by Euler's theorem, $13$ divides $a^{12}-a^0$, hence also $a^3(a^{12}-a^0)=a^{15}-a^3$.

Answer 2

Hint: Use Fermat or Euler theorem to prove that

$a^{15}-a^{3}$ is congruent to 0 mod 5, 7, 8, 9 and 13 respectively.

Answer 3

Hint $\ $ If $\,3\mid a\,$ then $\,3^2\mid a^3\mid a^3(a^{12}-1).\,$ Else $\,a\,$ is coprime to $\,3\,$ and $\,\phi(3^2) = \color{#c00}6,\,$ so by Euler, $\,{\rm mod}\ 3^2\!:\ a^{\color{#c00}{6}}\equiv1\,\overset{\rm square}\Rightarrow\,a^{12}\equiv 1^2\equiv 1.\,$ Therefore $\, 3^2\mid a^3(a^{12}-1)\,$ for all $\,a\in\Bbb Z.$

The same idea works for all other the other factors $\,p^n$ since $\,n\le 3\,$ so $\,p\mid a\Rightarrow p^n\mid a^3;$ otherwise $\,a\,$ is coprime to $\,p,$ so by Euler $\,p^n\mid a^{12}-1\,$ since $\,\phi(p^n)\mid 12\,$ in all cases. Therefore, generally

Theorem $ \!\!\!\!\! \underbrace{\,p_1^{n_1}\cdots p_j^{n_j}}_{\large p_i\, \rm distinct\ primes\ \ \ }\!\!\!\!\!\!\!\!\!\mid a^n(a^\phi-1)\ $ for all $\,a\,$ when $ $ all $\,n_i\le n\,$ and all $\,\phi(p^{n_i})\mid \phi\,$

See here for the simple proof, and links to many worked examples.

Answer 4

$a^{15} - a^3$ may be factored as $$a^{15} - a^3 = a^3(a^{12} - 1) = a^3(a^6 + 1)(a^6 - 1) = a^3(a^6 + 1)(a^3 + 1)(a^3 - 1) = a^3(a^6 + 1)(a^3 + 1) (a^2 + a + 1)(a - 1)$$ Replacing $a$ by 2, we get $$2^{15} - 2^3 = 2^3(2^6 + 1)(2^3 + 1)(2^2 + 2 + 1)(2 - 1) = 8 \times 65 \times 9 \times 7 \times 1 = 2^3. 3^2 . 5. 7 . 13$$ Now, $$\phi(8) = \phi(2^3) = 2^3 - 2^2 = 4$$ $$\phi(9) = \phi(3^2) = 3^2 - 3^1 = 6$$ $$\phi(5) = 5 - 1 = 4$$ $$\phi(7) = 7 - 1 = 6$$ $$\phi(13) = 13 - 1 = 12$$ There arise two possibilities:\ $\bullet$ If $\gcd(a, 2^{15} - 2^3) = 1$, then $$\gcd(a, 8) = \gcd(a, 9) = \gcd(a, 5) = \gcd(a, 7) = \gcd(a, 13) = 1$$ Thus, by Euler's theorem $$a^4 = a^{\phi(8)} \equiv 1 \mod 8 ~ \Rightarrow a^{12} \equiv 1 \mod 8$$ $$a^6 = a^{\phi(9)} \equiv 1 \mod 9 ~ \Rightarrow a^{12} \equiv 1 \mod 9$$ $$a^4 = a^{\phi(5)} \equiv 1 \mod 5 ~ \Rightarrow a^{12} \equiv 1 \mod 5$$ $$a^6 = a^{\phi(7)} \equiv 1 \mod 7 ~ \Rightarrow a^{12} \equiv 1 \mod 7$$ $$a^{12} = a^{\phi(13)} \equiv 1 \mod 13$$ Therefore, $$a^{12} \equiv 1 \mod(8 \times 9 \times 5 \times 7) \equiv 1 \mod (2^{15} - 2^3)$$ and hence $$a^{15} \equiv a^3 \mod (2^{15} - 2^3)$$ i.e. $2^{15} - 2^3$ divides $a^{15} - a^3$.\ $\bullet$ If $\gcd(a, 2^{15} - 2^3) \neq 1$, then we may write $a = k(2^{15} - 2^3)$ for some $k \in \mathbb{Z}$. Therefore, $$a^{15} - a^3 = a(a^{14} - a^2) = k(2^{15} - 2^3)(a^{14} - a^2)$$ and hence $$a^{15} \equiv a^3 \mod (2^{15} - 2^3)$$ i.e. $2^{15} - 2^3$ divides $a^{15} - a^3$.

Answer 5

Apply Carmichael function,

to find $\lambda(8)=2,\lambda(5)=\phi(5)=4, \lambda(9)=\phi(9)=6$ etc.

Then, for $9, F=a^{15}-a^3=a^3(a^{12}-1)=a^3(a^6-1)(a^6+1)$

As $3$ is prime, ether $(i)\ 3|a\implies 3^3|F$ or $(ii)\ (3,a)=1\implies 9|(a^6-1)$

For $8, F=a^{15}-a^3=a^3(a^{12}-1)$

As $2$ is prime, ether $(i)\ 2|a\implies 2^3|F$ or $(ii)\ (2,a)=1\implies 8|(a^2-1)$

For $5, F=a^{15}-a^3=a^3(a^{12}-1)=a^3\{(a^4)^3-1\}=a^3(a^4-1)(a^8+a^4+1)$

As $5$ is prime, ether $(i)\ 5|a\implies 5|F$ or $(ii)\ (5,a)=1\implies 5|(a^4-1)$

Similarly for $7,13$

Answer 6

We can factor: $$a^{15}-a^3=a^3(a-1)(a+1)(a^2+1)(a^2+a+1)(a^2-a+1)(a^4-a^2+1)$$

Now, we just need to prove that this is divisible by $2^3\cdot3^2\cdot5\cdot7\cdot13$.

If $a$ is odd, $(a-1)(a+1)(a^2+1)$ is the product of three even numbers, and is divisible by 8. If $a$ is even, $a^3$ is divisible by 8.

If $a$ is a multiple of 3, $a^3$ is divisible by $9$. Otherwise, one of $(a-1)$ and $(a+1)$ is divisible by 3, as is one of $(a^2-a+1)$ and $(a^2+a+1)$, and therefore $a^2(a-1)(a+1)(a^2-a+1)(a^2+a+1)$ is divisible by 9.

If $a$ is a multiple of 5, obviously so is our product. If not, $a^2$ is either 1 more or one less than a multiple of 5, so $a(a-1)(a+1)(a^2+1)$ is divisible by 5.

Consider $a\mod7$. If it's 0, then $a$ is divisible by 7. If it's 1, $(a-1)$ is divisible by 7. If it's 2 or 4, $(a^2+a+1)$ is divisible by 7. If it's 3 or 5, $(a^2-a+1)$ is divisible by 7. If it's 6, $(a+1)$ is divisible by 7. Regardless, our product is divisible by 7.

Consider $a\mod13$. If it's 0, then $a$ is divisible by 13. If it's 1, $(a-1)$ is divisible by 13. If it's 2 or 6 or 7 or 11, $(a^4-a^2+1)$ is divisible by 13. If it's 3 or 9, $(a^2+a+1)$ is divisible by 13. If it's 4 or 10, $(a^2-a+1)$ is divisible by 13. If it's 5 or 8, $(a^2+1)$ is divisible by 13. If it's 12, $(a+1)$ is divisible by 13. Regardless, our product is divisible by 13, and we're done.

Answer 7

Use the hint. Show that 5,7,8,9 and 13 divide any $a^{15}-a^3$. Also, the factorization of $a^{15}-a^3$ is $a^3(a^{12}-1)$. If $a$ is divisible by any member of $\{5,7,8,9,13\}$, then $a^{15}-a^3$ is also divisible by that number. If it's not, then by Fermat little theorem $$a^p-1 \equiv 0\mod p$$ for the primes in the set. Then show that $a^p-1$ divides $a^{12}-1$. Now for $8$ and $9$, you'll have to go the more traditional way, but they're easy to show.

Answer 8

This is meant mostly as a joke, but partly not:

We clearly have $2^{15}-2^3=32{,}760$ divides $a^{15}-a^3$ when $a=0$, $\pm1$, and $\pm2$, so that leaves only $32{,}755$ other cases to check....

(The part that's not meant as a joke is the implicit question, where do you draw the line between proving something by brute force computation versus spending time looking for a clever theory.)