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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.

Basem Fouda
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Write it as $$\frac{3y-1}{y-3}=\frac { 3\left( y-3 \right) +8 }{ y-3 } =3+\frac { 8 }{ y-3 } $$

haqnatural
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write $$\frac{3y-1}{y-3}=\frac{3y-9+8}{y-3}=3+\frac{8}{y-3}$$

Dr. Sonnhard Graubner
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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.

S.C.B.
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  • can you please continue, I am not too familiar with using mod but I get the general idea. – Basem Fouda Jan 29 '17 at 15:46
  • by not so much do you mean not? – Asinomás Jan 29 '17 at 15:54
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    for example: Michael Jordan was really good at basketball and not so much at baseball, but he was still good at it. – Asinomás Jan 29 '17 at 15:55
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    @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. – S.C.B. Jan 29 '17 at 15:57
  • obviously, every interval of time contains an uncountably infinite number of moments. – Asinomás Jan 29 '17 at 16:02
  • When 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. – Bill Dubuque Jan 29 '17 at 17:28
  • @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. – S.C.B. Jan 30 '17 at 00:17
  • @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 $y-3= \pm 1, \pm 2, \pm 4, \pm 8$. – S.C.B. Jan 30 '17 at 00:18
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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 $$y \in \left\{\frac{8}{k} + 3\ |\ k \in \{1,-1,2,-2,4,-4,-8,8\}\right\}$$

Maciej Caputa
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  • Your 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* – Bill Dubuque Jan 29 '17 at 17:39
  • Could 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.

     – Maciej Caputa Jan 29 '17 at 18:05
  • We seek integral $y,$ so in the final equation $k$ must be restricted to divisors of $8$, not "all $k\in\Bbb Z,\ldots$ – Bill Dubuque Jan 29 '17 at 18:31
  • Please 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

    $$\frac{8}{y - 3} + 3 = \frac{8}{ \frac{8}{7} + 3 - 3} + 3 = \frac{8}{\frac{8}{7}} + 3 = \frac{8}{1} \times \frac{7}{8} + 3 = 7 + 3 = 10$$

     – Maciej Caputa Jan 29 '17 at 18:36
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    It 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 $k$" – Bill Dubuque Jan 29 '17 at 18:53
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    Oh... Thank you for your patience in explaining it. Now I understand your point. I will correct my answer. – Maciej Caputa Jan 29 '17 at 20:00
  • Great! $\phantom{................}$ – Bill Dubuque Jan 29 '17 at 20:03
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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.

H. H. Rugh
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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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Bill Dubuque
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