Basically, I have to use Euler theorem to prove that $10n^8 -9n^6 -n^2$ is divisible by $45$
So my approach so far is to say
$10n^8 - 9n^6 - n^2 = 0 \bmod 45$
Now $45$ can be factored into $5$ and $9$
So I figure I need to do something like
$10n^8 - 9n^6 - n^2 \bmod 5$
$10n^8 - 9n^6 - n^2 \bmod 9$
But I am not sure how to exactly prove it.
I know how to prove something like $n^{13} - n$ is a multiple of $2730$ using Fermat's theorem, but the proof works nicely because the exponent is a prime. My approach to $n^{13} - n$ would consist of proving that $n^{13} - n$ is divisible by $\{2, 3, 5, 7, 9, 13\}$.
What is throwing me off is the coefficients in $10n^8 - 9n^6 - n^2$ and the fact that the exponents are not primes.
$\endgroup$ 52 Answers
$\begingroup$To prove $10n^8 - 9n^6 - n^2\equiv0\mod5$, it is enough to show $n^6-n^2\equiv0\mod5$. This follows from $n^5\equiv n\mod5$.
To prove $10n^8 - 9n^6 - n^2\equiv0\mod9$, it is enough to show $n^8-n^2\equiv0\mod9$. This follows from $n^7\equiv n\mod9$, since $\phi(9)=6$.
$\endgroup$ 2 $\begingroup$HINT
$$10n^8 - 9n^6 - n^2\equiv 0n^8-(-1)n^6 - n^2 \equiv n^6 - n^2 \pmod 5$$
By little fermat we have $n^5\equiv n \pmod 5$, therefore $$n^6-n^2\equiv n\cdot n^5-n^2\equiv n\cdot n-n^2\equiv n^2-n^2\equiv 0\pmod{5}$$
See if you can work the other mod similarly
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