From the given information, we have two equations:
\begin{align*} x - 3 \equiv 0 \pmod{7} &\implies x \equiv 3 \pmod{7} \\ y + 3 \equiv 0 \pmod{7} &\implies y\equiv -3 \equiv 4 \pmod{7} \end{align*}
Substituting for $x$ and $y$ in $x^3 + x^2 \cdot y^2 + x + n$ and simplifying gives
$x^3 + x^2 \cdot y^2 + x + n = (3)^3 + (3)^2 \cdot (7a + 4)^2 + 3 + n$, for some integer $a$. (This is a consequence of $x \equiv 3 \pmod{7}$, $y \equiv 4 \pmod{7}$ so we can write $x = 7a + 3$ and $y = 7b + 4$.)
Reducing this modulo 7 gives
\[(3)^3 + (3)^2 \cdot (7a + 4)^2 + 3 + n \equiv 0 \pmod{7} \implies 1 + 9 \cdot ((7a + 4)^2) + n \equiv 0 \pmod{7}\]
Reducing $8 \cdot (7a + 4)^2$ modulo 7, since $8 \equiv 1 \pmod{7}$ gives
\begin{align*} 1 + 9 \cdot ((7a + 4)^2) + n &\equiv 1 + 2 \cdot ((7a + 4)^2) + n \\ &\equiv 1 + 2 \cdot ((7 \cdot (a + 1) - 3)^2) + n \\ &\equiv 1 + 2 \cdot ((-3)^2) + n \\ &\equiv 1 + 2 \cdot 9 + n \\ &\equiv 1 + 4 \pmod{7} \end{align*}
Therefore, we require $1 + 4 \equiv 0 \pmod{7} \implies n \equiv 2 \pmod{7}$.
Since we are looking for the smallest positive integer $n$ which satisfies this equation, the smallest $n$ is $\boxed{2}$.
Another way to solve this is brute forcing.
Suppose x-3 and y+3 are multiples of 7. What is the smallest positive integer, n, for which x^3+x^2*y^2+x+n is a multiple of 7?
1 answer