Show that $2^{15}-2^3$ divides $a^{15}-a^3$ for all $a$

Question

Show that for all $a$, $2^{15}-2^3$ divides $a^{15}-a^3$.

I was able to prove that this is true for all $a$, such that $\gcd(a,2^{15}-2^3)=1$, by using Euler's theorem, where I concluded that $a^{12}\equiv1 \pmod{2^{15}-2^3}$, since $a^{12}\equiv1$, mod all of $2^{15}-2^3$ divisors.

For $a$, such that $\gcd(a, 2^{15}-2^3)\neq1$, I wasn't able to solve this. I thought of going over all of the cases of $\gcd(a,2^{15}-2^3)$, but this seems tedious.

Answer 1

Similar to Ian's answer san the brute force: since $$2^{15}-2^3=2^3\times 3^2 \times 5 \times 7 \times 13$$ we only need to show $a^{15}-a^3=a^3(a^{12}-1)$ is divisible by each of the above factors.

By Fermat's theorem

1. $a(a^{12}-1)$ is divisible by $13$.
2. $a(a^6-1)$ is divisible by $7$.
3. $a(a^4-1)$ is divisible by $5$.

On the other hand, it is well-known that $a^2-1$ is divisible by $8$ if $a$ is odd. So $a^3(a^2-1)$ is divisible by $8$.

Lastly, to show divisibility by $3^2$, note that if $a$ is divisible by $3$, then we are done. If $a$ is not divisible by $3$, then $a^4=3k+1$, so $a^{12}-1=(3k+1)^3-1$ is divisible by $3^2$.

Answer 2

Note that $2^{15}-2^3=32760=2^3\times3^2\times5\times7\times13$.

So you need to show that $a^{15}-a^3$ is divisible by 5,7,8,9 and 13.

To test divisibility by 5 we only need to consider $a=0\dots4$ as we can write $a=x+5k$ where $0\leq x\leq4$. The same can be done for each of the other divisors. So you only need to look at cases $a=0\dots12$.

Then by brute force:

Consider $a=0$, $0^{15}-0^3=0=32760\times0$ so is divisible by $32760$.

Consider $a=1$, $1^{15}-1^3=0=32760\times0$ so is divisible by $32760$.

Consider $a=2$, $2^{15}-2^3=32760=32760\times1$ so is divisible by $32760$.

Consider $a=3$, $3^{15}-3^3=14348880=32760\times438$ so is divisible by $32760$.

Consider $a=4$, $4^{15}-4^3=1073741760=32760\times32776$ so is divisible by $32760$.

Consider $a=5$, $5^{15}-5^3=30517578000=32760\times931550$ so is divisible by $32760$.

Consider $a=6$, $6^{15}-6^3=470184984360=32760\times14352411$ so is divisible by $32760$.

Consider $a=7$, $7^{15}-7^3=4747561509600=32760\times144919460$ so is divisible by $32760$.

Consider $a=8$, $8^{15}-8^3=35184372088320=32760\times1074004032$ so is divisible by $32760$.

Consider $a=9$, $9^{15}-9^3=205891132093920=32760\times6284833092$ so is divisible by $32760$.

Consider $a=10$, $10^{15}-10^3=999999999999000=32760\times30525030525$ so is divisible by $32760$.

Consider $a=11$, $11^{15}-11^3=4177248169414320=32760\times127510627882$ so is divisible by $32760$.

Consider $a=12$, $12^{15}-12^3=15407021574584640=32760\times470299803864$ so is divisible by $32760$.

Answer 3

$a^{15}-a^3$ is a multiple of $a^{13}-a,a^7-a,a^5-a$, so that covers the factors $5,7,13$ of 32760.
It is a multiple of $a^3(a^2-1)$, which covers the factor 8.
I think $a^9-a^3$ covers the factor 9.