I'm sure there must be an easier solution, but here's mine: \[ k(k^{2}+3ka+3a^{2})\equiv 0\pmod{2007}.\] Looking modulo 3, we observe that $ k\equiv 0\pmod 3$. Therefore let $ k = 3x$, $ x$ being a positive integer. We can prove now that all multiples of 3 are good. Check http://www.mathlinks.ro/Forum/viewtopic.php?p=934678#934678

A small addendum/correction to Valentin's nice solution: There is some confusion with k being divisible by 3 and k being divisible by 9: Valentin correctly shows that k=3x, and that x must be a multiple of 223. Hence, k must be a multiple of 669. (That's the gist if his argument.) Every occurrence of "2007" after the second line of Valentin's proof should be replaced by "669". (a repeated typo).

$ k^{2}+3ka+3a^{2}\equiv 0(\mod 223)$ You can use algebra number $ x\equiv (\frac{a}{k})(\mod 223)$ $ 3x^{2}+3x+1\equiv 0(\mod 223)$ Use quadric reside.

test20 wrote: Every occurrence of "2007" after the second line of Valentin's proof should be replaced by "669". (a repeated typo). Oops, I was in a hurry. Obviously $ 3\times 223 = 669\neq 2007$

Let's put a=38t and k=3t.... (a+k)^3-a^3=(41t)^3-(38t)^3=(41^3-38^3)t^3=(68921-54872)t^3=14049*t^3=7*2007*t^3. The result is multiple of 2007;;; Am I wrong??

You are right limsk1, your answer is right and your solution is very nice. Try a more general problem, with a natural number b instead of 2007 and natural number c instead of 3. I mean for given numbers b and c, find all numbers k, such that there exist a number a, that: $ a^{c}\equiv (a+k)^{c}(mod b)$

The Polynomial congruence quite hard to solve $ f(x)\equiv 0(mod p)$ $ x = (\frac{k}{a})(\mod b)$ You can find out all b such that exist $ (a,k)$ and $ \gcd(a,b) = 1$ $ (a+k)^{k}-a^{3}\equiv 0(\mod b)$

It seems that I did a computational mistake in my problem, as the congruence is solvable Thanks for pointing that out limsk1

Another solution: Proposition: Let $ p\geq 3$ be a prime number and let $ d > 1$ be a divisor of $ p - 1$. Then there exists an integer $ a$ so that $ (a + k)^d\equiv a^d\pmod p$. Proof: if $ p\mid k$ then it's obvious so suppose that $ p\nmid k$. Let us furthermore assume that $ p\nmid a$. Then notice that the congruence is equivalent to \[ \left(1 + \frac {k}{a}\right)^d \equiv 1 \pmod p \] (if anyone is not familiar with fractions in modular arithmetic - there are enough posts about it on ML). Let $ g$ be a primitive root modulo $ p$. Then we see that \[ a\equiv \frac {k}{g^{\frac {p - 1}{d}} - 1} \pmod p \] is a solution since \[ \left(1 + \frac {k}{a}\right)^d \equiv \left(1 + \frac {k\cdot (g^{\frac {p - 1}{d}} - 1)}{k}\right)^d \\ \equiv g^{p - 1} \equiv 1 \pmod p. \] $ \Box$ It's obvious that $ 3\mid k$ is a neccessary and sufficient condition for divisibility by $ 9$, so from the above, we see that the solutions are the multiples of $ 3$.

it suffices to consider mod 9,223. by$9|(a+k)^3-a^3$ we get $3|k^3$ so$3|k$ easy to check it's sufficient. when mod 223,if $223|k$ it's trivial if $(k,223)=1$ let g be an original root mod 223 then$a+k\equiv ag^{74} or ag^{148}$ it's easy to calculate {g^74,g^148}={39,183} hence $k=38a or 182a(mod 223)$ since$ (38,223)=(182,223)=1$ so for any resicue class mod 223,we can choose an appropriate a satisfying these conditions. so $3|a$ is the answer.