I have encountered this as part of a bigger problem but I really don't know how to go on about it. I would also appreciate it if you could specify a certain technique to follow when facing such a problem.
For what integer values of $y$ is $\frac{3y-1}{y-3}$ an integer?
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$\begingroup$
elementary-number-theory
prime-factorization
integers
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0The answer you accepted is incorrect - see my comments on it. – 2017-01-29
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0The answer has been corrected. – 2017-01-31
6 Answers
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Hint $\ y\!-\!3\mid f(y)\iff y\!-\!3\mid f(3)\ $ for any polynomial $f$ with integer coefficients, because
$\ {\rm mod}\,\ y\!-\!3\!:\ y\equiv 3\,\Rightarrow\, f(y)\equiv f(3)\ $ by the Polynomial Congruence Rule.
Equivalently $\, f(y)\equiv f(n)\pmod{y\!-\!n},\ $ the Polynomial Remainder Theorem.
Alternatively we can apply the Euclidean algorithm and the remainder theorem as follows
$$\gcd(y\!-\!3,f(y))\, =\, \gcd(y\!-\!3,\,f(y)\bmod y\!-\!3)\, =\, \gcd(y\!-\!3,f(3))$$
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Write it as $$\frac{3y-1}{y-3}=\frac { 3\left( y-3 \right) +8 }{ y-3 } =3+\frac { 8 }{ y-3 } $$
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0Wat would be the general formula for $y$ then and how to find it, sorry if my question is too easy. – 2017-01-29
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1It depends on the problem, just try always factor with respect to the denominator – 2017-01-29
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0thank you this helped me very much. – 2017-01-29
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write $$\frac{3y-1}{y-3}=\frac{3y-9+8}{y-3}=3+\frac{8}{y-3}$$
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2What does this add to @haqnatural's answer? – 2017-01-29
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7i have written this without knowing the other answers – 2017-01-29
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4@S.C.B. There was a gap of only seconds between haqnatural's post, and the Dr.'s post, certainly not enough time to copy another's post. This deserves as much credit as the one you speak of. harassment. – 2017-01-29
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1oh man is this a challenge here? – 2017-01-29
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1@amWhy Sorry, there seemed to be a larger time gap to me. I wasn't trying to harass him – 2017-01-29
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0I am sorry; I may not have been thinking clearly. It is time to sleep where I am at. – 2017-01-29
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0O.o it's 1 am in Korea. – 2017-01-29
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The technique is by using modulos. $$3y-1=3(y-3)+8 \equiv 0 \pmod {y-3}$$ So $8 \equiv 0 \pmod{y-3}$. I think you can continue from here.
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0can you please continue, I am not too familiar with using mod but I get the general idea. – 2017-01-29
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0by not so much do you mean not? – 2017-01-29
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1for example: Michael Jordan was really good at basketball and not so much at baseball, but he was still good at it. – 2017-01-29
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1@Jorge Please, I do not think that this is really suited in my answer. You should take this to the chat room and discuss it there. This kind of comments are not recommended, if there are are they might get reported. – 2017-01-29
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0obviously, every interval of time contains an uncountably infinite number of moments. – 2017-01-29
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0When you use modular methods you should *not* compute/dsplay the obfuscatory/unneeded quotient - that defeats the entire purpose, e.g. see my answer. The quotient plays no role - only the remainder. – 2017-01-29
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0@BillDubuque Good point. I first didn't write the quotient. But then I thought that it would be a bit unclear for the OP, so I editted. But then I realized I didn't need modulos then. However, it was too late to edit-Dr. Sonnhard and haqnatural had already added their solutions. So I'm just going to keep it as it is for now. – 2017-01-30
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0@BasemFouda What I'm doing here is the basically nearly the same thing as Dr. Sonnhard and haqnatural. See Billdubque. Since $3y-1$ is the same as $8$ modulo $y-3$, we can conclude that $8$ is $0$ modulo $y-3$, which would imply that $y-3$ is a divisor of $8$. This gives us that $y-3= \pm 1, \pm 2, \pm 4, \pm 8$. – 2017-01-30
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Please, note that we encounter here with a rational function of the form $x = f(y) = \frac{g(y)}{h(y)}$ where $g(y) = 3y - 1$ and $h(y) = y - 3$ with a restriction that $h(y) \neq 0 \iff y \neq 3$.
For any rational function it is possible to express $f(y)$ in a form of $\frac{a}{y - p} + q$ where $a, p, q \in \mathbb{R}\ \wedge\ a, p, q= \textrm{const}$ and lines $y = p$ and $x = q$ are the asymptots of the function $f(x)$.
Hence, $$f(y) = \frac{3y - 1}{y - 3} = \frac{3(y - 3) + 8}{y - 3} = \frac{8}{y - 3} + 3$$
This yields, that $$f(y) \in \mathbb{Z} \iff y - 3\ \textrm{mod}\ 8 \equiv 0$$
In other words, $f(y)$ is an integer if and only if $8$ is divisible by $y-3$, which implies that $8 = k(y - 3)$ for some $k \in \mathbb{Z}$ ($8$ is a multiple of $y-3$).
Finally,
$$8 = k(y - 3)$$
$$y = \frac{8}{k} + 3$$
This is satisfied for all $k \in \{z \in \mathbb{Z}\backslash \{0\}\ |\ 8\ \textrm{mod}\ z \equiv 0\} = \{1,-1,2,-2,4,-4,-8,8\}$ since $y$ must be an integer.
Therefore solution set can be given as $$y \in \left\{\frac{8}{k} + 3\ |\ k \in \{1,-1,2,-2,4,-4,-8,8\}\right\}$$
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0Your final equiation is incorrect - it is not true for *all* $k\in \Bbb Z.\ $ The remarks about rational functions are unneeded - this is a problem in *number theory** – 2017-01-29
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0Could you elaborate on why is the final equation incorrect and give a counterexample of value of $k \in \mathbb{Z}\ \wedge\ k \neq 0$ for which $y$ given as $f(\frac{8}{k} + 3)$ doesn't yield an integer number? Author of the question asks for guidance on approaching such problem, therefore I have outlined a procedure of expressing rational function in an asymptotic form. – 2017-01-29
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0We seek *integral* $y,$ so in the final equation $k$ must be restricted to divisors of $8$, not "all $k\in\Bbb Z\,\ldots$ – 2017-01-29
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0Please re-read my answer. Here is a counter example of your statement. Let's take $k = 7$, nota bene $7$ is **not** a divisor of $8$. Hence we get $y = \frac{8}{k} + 3 = \frac{8}{7} + 3$ Substituting into the original equation we get $$\frac{8}{y - 3} + 3 = \frac{8}{ \frac{8}{7} + 3 - 3} + 3 = \frac{8}{\frac{8}{7}} + 3 = \frac{8}{1} \times \frac{7}{8} + 3 = 7 + 3 = 10$$ – 2017-01-29
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2It seems you misunderstood my prior comment. Look at the question title. Note that it reads "For what *integer* values of $y$ ...". So the solution $y$ must be an *integer.* But $\, y=8/k + 3\,$ is an integer only when $k$ divides $8$, not as you claim "for all nonzero integers $k$" – 2017-01-29
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1Oh... Thank you for your patience in explaining it. Now I understand your point. I will correct my answer. – 2017-01-29
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0Great! $\phantom{................}$ – 2017-01-29
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hint: Suppose it equals an integer $n$ and rewrite it as $(y-3)(n-3)=8$. This leaves few possibilities to go through.