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The following exercise is from [Birkhoff and MacLane, A Survey of Modern Algebra]:

For which rational numbers $x$ is $3x^2-7x$ an integer? Find necessary and sufficient conditions.

I think I was able to obtain the set of rationals $x$, but I am not sure what the necessary and sufficient conditions are about. Here is what I tried: Suppose $3x^2-7x=k, k \in \mathbb{Z}$. Solving the quadratic, we get $x=\frac{7 \pm \sqrt{49+12k}}{6}$, whence $49+12k$ must be the square of some integer $m$. Now $m^2 \equiv 49 (\operatorname{mod} 12)$ has solutions $m = 1,5,7,11 (\operatorname{mod} 12)$, i.e. $m \in \{1,5,7,11\}+12 \mathbb{Z}$. Thus, the set of rationals $x = \frac{7 \pm m}{6}$ is $\{0,\frac{1}{3} \}+\mathbb{Z} = \{\ldots,0, \frac{1}{3}, 1, \frac{4}{3}, \ldots \}$. Is this correct? What do the necessary and sufficient conditions refer to?

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What you've found are called "necessary conditions" on $x$ for $3x^2-7x$ to be an integer--so called, because if the conditions fail, then $3x^2-7x$ will not be an integer. "Sufficient conditions" are conditions on $x$ that that, if they hold, will imply that $3x^2-7x$ is an integer. Your conditions on $x$ are also sufficient conditions, as you can check. When we say "necessary and sufficient conditions," we mean equivalent conditions, or exact conditions.

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  • $\begingroup$ I am aware of what a necessary condition (and sufficient condition) is. I had found the exact set whereas I thought the exercise was asking to find a condition that is just necessary and one that is just sufficient. I guess the exercise asked to find a condition which is both necessary and sufficient. $\endgroup$
    – AG.
    Jul 12, 2013 at 17:44
  • $\begingroup$ I can empathize! I thought that "necessary and sufficient" meant the same thing as you thought, when I first started out. The key word there is "and," though. $\endgroup$ Jul 12, 2013 at 22:27
  • $\begingroup$ well, the word "conditions" is plural, which got me to think there was more the exercise was asking than just one exact condition that is both necessary and sufficient. $\endgroup$
    – AG.
    Jul 12, 2013 at 23:48
  • $\begingroup$ Those were my thoughts, exactly. Ambiguous phrasing. Well, we know better, now. $\endgroup$ Jul 13, 2013 at 2:14
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Your solution is correct, but here is a perhaps simpler one.

Write $x=u/v$ with $u$, $v$ coprime integers and $v>0$. Then $3x^2-7x$ is an integer iff $v^2$ divides $3u^2-7uv$. This implies that $v$ divides $3u^2$. Since $u$ and $v$ are coprime, we get that $v$ divides $3$, that is, $v=1$ or $v=3$.

When $v=1$, $x$ is an integer, no surprises there.

When $v=3$, we get that $9$ divides $3u^2-21u$ and so $3$ divides $u^2-u=u(u-1)$. Since $u$ and $v=3$ are coprime, we get that $3$ divides $u-1$, that is, $u=1+3t$ and $x=t+1/3$, with $t$ an arbitrary integer.

This coincides with your solution, of course.

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