Find all integers n such that number $n^2-5n+6$ is divided number $n-3$

Question

$n-3|n^2-5n+6$

Note that $n^2-5n+6=(n-3)^2-(n+3)$

$n-3$ divides $(n-3)^2$, then $n-3$ divides $n+3$ too.

Then $n+3=0 -> n=-3$

If $n=-3$ then $n^2-5n+6=9+15+6=30$

Number $30$ divides +-$\{ 1, 2, 3, 5, 6, 10, 15, 30\}$

Then $n\in \{-33,-18,-13,-9,-8,-5,-4,-2,-1,0,2,3,7,12,27\}$

Is this correct? Because if n=1, it's correct to.

Thank You.

Answer 1

I was not going to write an answer, but none of the above contains the final statement.

Since $n^2-5n+6=(n-2)(n-3)$, the number $n-3$ is a divisor of $n^2-5n+6$ in all cases in which it is nonzero.

So the final answer is: $$\mathbb{Z}\setminus\{3\}$$, all integers except $3$ (for $n=3$ you have $n-3=0$, this can't be a divisor!).

Answer 2

note that $$n^2-5n+6=(n-2)(n-3)$$

Answer 3

Notice that $n^2-5+6 = (n-2)(n-3)$ for all integers $n$, so $n-3|n^2-5+6$ all ways.

Answer 4

If $n^2 - 5n + 6 = x$ then $x = (n-2)(n-3)$, follows that $n-3\mid n^2 - 5n + 6$.

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Proof: We want to factor $x$, meaning to let $x = (n + y_1)(n + y_2)$ for constants $y_1$ and $y_2$.

This means that $$y_1y_2 = 6$$ and $$y_1 + y_2 = -5.$$ The answer is obtained by letting $y_1 = -2$ and $y_2 = -3$ since $$(-2)(-3) = 2\times 3 = 6$$ and $$(-2)+(-3) = -2 - 3 = -5.$$ Therefore, $$\begin{align} x &= (n + -2)(n + -3) \\ \Leftrightarrow n^2 - 5n + 6 &= (n-2)(n-3).\end{align}$$ And thus, we get that $$n-3\mid n^2 - 5n + 6$$ as desired. $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,\Box$

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For a simpler approach, the following is going to be exactly what @DietrichBurde mentioned, but clearer so you know precisely where your mistake is. $$(n-3)^2 - (n+3) = n^2 + 3^2 - 2\cdot 3\cdot n - n - 3 = n^2 + 9 - 6n - n - 3 = n^2 - 7n - 6.$$ This yields the wrong equation, therefore we need to $+n$ instead of $-n$, which implies that $$n^2 - 5n + 6 = (n-3)^2 + (n - 3) = (n-3)(n-3 + 1) = \boxed{ \ (n-3)(n-2). \ }\tag*{$\Box$}$$