<s>[INST] Find all $n$ natural numbers such that $10\mid n^{10}+1$ Determine all natural numbers $n$ such that : 
$10$ divisor of $n^{10}+1$ 
My attempt : 
Let $n=r(\mod{10})$ so $n^{10}+1=(r^{10}+1)(\mod{10})$ 
This mean : $r^{10}+1=0(\mod{10})$ 
Now $r\in {0,1,2,3,4,5,6,7,8,9}$ after try I get $r=3,7$ 
So : $n=10k+3,10k+7$ 
Is my work correct ? 
Please I need other simple method to computing 
 [/INST] You are correct but a few things to make you calculations much fewer.
1) if $n$ and $10$ are not relatively prime then any common divisor $a;a\ne 1$ will not divide $n^{10}+1$ so $10\not  \mid n^{10}+1$.
So we need only test $1,3,7,9$
2) $(10-i)^{10}\equiv i^{10}\pmod{10}$ so we only check $1,3$.
3) Eulers Thereom says as $\phi(10) =4$ that for $n$ relatively prime to $10$ that $n^4\equiv 1 \pmod {10}$ so $n^{10}\equiv n^2 \pmod {10}$
So we just need to check whether $k^2 + 1\equiv 0 \pmod {10}$ for $k= 1,3$.
We get $k=3$.
and so our solutions are $n\equiv \pm 3 \pmod {10}$ 
or we could say $n\equiv 3,7 \pmod {10}$
or we could say $n = 10k +3$ or $n=10k +7$ for some integer $7$
or we could say "$n$ is any number with the last digit of $3$ or $7$".
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