$2005|(a^3+b^3) , 2005|(a^4+b^4 ) \implies2005|a^5+b^5$ How can I show that if $$2005|a^3+b^3 , 2005|a^4+b^4$$ then $$2005|a^5+b^5$$
I'm trying to solve them from $a^{2k+1} + b^{2k+1}=...$ but I'm not getting anywhere.
Can you please point in me the correct direction?
Thanks in advance
 A: $(a^5+b^5) = (a+b)(a^4+b^4) - ab(a^3+b^3)$.
So for any $n$, if $n \mid a^3+b^3$ and $n \mid a^4+b^4$ then $n \mid a^5+b^5$.
A: Actually you don't need the $a^4 + b^4$.
$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. It turns out that $a^2 - ab + b^2 \equiv 0$ has no solutions except $(0,0)$ mod either $5$ or $401$, so the only way to have $a^3 + b^3 \equiv 0 \mod 2005$ is 
$a + b \equiv 0$, and then you also have $a^k + b^k \equiv 0 \mod 2005$ for every odd positive integer $k$.
A: If prime $p|(a^3+b^3), p|(a+b)(a^3+b^3) \implies p| \{(a^4+b^4)+ab(a^2+b^2)\}$
If $p|(a^4+b^4), p$  must divide $ab(a^2+b^2)$
If $p|a,$ $p$ must divide $b$ as $p|(a^3+b^3)$ 
If $p|a$ and $p|b,$ then $p|(a^n+b^n)$ for integer $n\ge1$
Else $p\not\mid ab $  and  $p|(a^2+b^2)\implies p|(a+b)(a^2+b^2) \implies p| \{(a^3+b^3)+ab(a+b)\}$
As $p|(a^3+b^3),p|ab(a+b)\implies p|(a+b)$ as $p\not\mid ab$
As $p|(a+b)$ then $p|(a^m+b^m)$ for odd integer $m\ge1$
Now put $p=5,p=401$ separately and use lcm$[5,401]=2005$ 
A: Unless I made a mistake, we have a prime factorization $2005=5\cdot401$.
Idea: Show divisibility by $5$ and $401$ separately.
Hint #1: Little Fermat modulo $5$.
Hint #2: $\gcd(400,3)=1$, so $x\mapsto x^3$ is an automorphism of $\mathbb{Z}_{401}^*$.
A: a triffle answer;we have $2005=401.5$ so $p=5,401$ also: $$a^5+b^5=(a+b)(a^4+b^4-ab(a^2-ab+b^2)) \tag I$$.also we have 
$$a^3+b^3=(a+b)(a^2-ab+b^2) \tag {II} $$  from the $p|a^3+b^3$ we have 2 cases :  1) if $p|a+b$ then  from $(I)$ we have $p|a^5+b^5$
2)if $p|(a^2-ab+b^2)$ by attention to $(II)$,$p|(-ab)(a^2-ab+b^2)$and therefore $p|a^5+b^5$ by(I) and $p|a^4+b^4$.$$$$
 next subsequent step is using definition of LCD:$2005=[401,5]|a^5+b^5$ .
