$\begin{aligned} f(x) &= \frac{3x+2018}{2x+2017} \\ &= \frac{3x+3025.5}{2x+2017} - \frac{1007.5}{2x+2017} \\ &= \frac{3}{2}\bigg(\frac{x+1008.5}{x+1008.5}\bigg) - \frac{1}{2}\bigg(\frac{2015}{2x+2017}\bigg)\\ &= \frac{3}{2} - \frac{1}{2}\bigg(\frac{2015}{2x+2017}\bigg)\\ \end{aligned}$

$f(x)$ is an integer $\Longrightarrow$ $\large \frac{2015}{2x+2017}$ is an odd number $\Longrightarrow$ $2x+2017$ must be the factor of 2015.

$2015 = 5^{1} \times 13^{1} \times 31^{1}$

From the above, we found that the factors of 2015 are $\pm1, \pm5, \pm13, \pm31, \pm65, \pm155, \pm403, \pm2015$ , total $16$ factors.

Recall: $2x+2017$ must be the factor or 2015.

Hence, the number of integers $x$ equal to the number of the factors of 2015 which is $\boxed{16}$ . $$ [Quick Solution —— Prime Number Decomposition Theorem]

Hence, the number of integers $x = 2 \times (1+1) \times (1+1) \times (1+1) = \boxed{16}.$

Wiki: https://www.cut-the-knot.org/blue/NumberOfFactors.shtml

$f(x)=\frac{3x+2018}{2x+2017}$ Multiply both sides by $2$ , and rearrange the top of the fraction to match the divisor: $2\times f(x)=\frac{6x+4036}{2x+2017}=\frac{6x+6051-2015}{2x+2017}=\frac{6x+6051}{2x+2017}-\frac{2015}{2x+2017}=3-\frac{2015}{2x+2017}$ Which can be rearranged to: $2015=(3-2\times f(x))\times (2x+2017)$ As $f(x)$ and $x$ are both integers, $3-2\times f(x)$ and $2x+2017$ must be cofactors of $2015$ , and also must be odd. Since all factors of $2015$ are odd, they are all viable solutions, and so the number of possible solutions is equal to the number of cofactor pairs of $2015$ , which is $\boxed{16}$ .

Brute force solution using PHP to yield 16 factors:

Rewrite $f(x)=\frac{3x+2018}{2x+2017}=\frac{x+1}{2x+2017}+1$ , for this to be an integer, $\frac{x+1}{2x+2017}$ has to be an integer.

Let $\frac{x+1}{2x+2017}=k$ where $k$ is an integer, we have, by Simon's Favorite Factoring Trick :

$x+1=2xk+2017k$ $2xk-x+2017k-1=0$ $4xk-2x+4034k-2=0$ $(2x+2017)(2k-1)=-2015$

Now since $2015=5\times13\times31$ , it has $2\times2\times2=8$ factors, that means if we include the negatives, $-2015$ will have $16$ factors. Every factor is odd, which matches the parity of the terms $(2x+2017)$ and $(2k-1)$ , which are both also odd. Therefore the number of integers $x$ such that $f(x)$ is an integer equals to the number of factorizations of $-2015$ , which is $16$ .

If $2n+2017$ divides $3n+2018$ , then it divides also $3 \cdot(2017+2n) -2 \cdot (3n+2018)$ ) which is $2015$ . The expression $2017+2n$ ranges over the odd integers (try plugging in $n=-1,n=-2,...$ and you will obtain $2015, 2013,...$ .

Since $2015=5 \cdot 13 \cdot 31$ , it has a total number of $2 \cdot 2\cdot2=8$ odd positive divisors. Now you can choose the sign of each positive divisor (if $65$ divides $2015$ so does $-65$ ) leading to $2 \cdot 8 =16$ possibilities, which can all be realized varying $n$ .

If the last part is unclear, try constructing a divisor of $2015$ from its prime factors: you can choose if you want to include $5$ or not (two options) and the same goes for other factors (when you choose none, you obtain $5^0\cdot13^0\cdot31^0=1$ .

Generalizing this reasoning, a number whose prime factorization is $p_1^{a_1} \cdot p_2^{a_2 }\cdot \dots \cdot p_m^{a_m}$ has $(a_1+1)\cdot\dots\cdot(a_m+1)$ positive divisors.

We need to have $\frac{2018 + 3n}{2017 + 2n} = k$ for some integer $k$ . Multiply out the denominator:

$2018 + 3n = 2017 k + 2 n k$ ,

and group all the terms together on one side and set the other side equal to zero:

$2017 k + 2 n k - 3 n - 2018 = 0$ .

Dividing everything by $2$ gives:

$n k - \frac{3}{2} n + \frac{2017}{2} k -1009 = 0$ .

This almost looks like the setup for Simon's Favorite Factoring Trick . In this case we would have $j = \frac{2017}{2}, i = - \frac{3}{2}$ , but the problem is $-1009 \not= i j = -\frac{6051}{4}$ . To get this expression into the desired form, add $-\frac{6051}{4}$ to both sides and apply the factoring trick:

$(k - \frac{3}{2})(n + \frac{2017}{2}) - 1009 = -\frac{6051}{4}$ ,

$(k-\frac{3}{2})(n + \frac{2017}{2}) = 1009 - \frac{6051}{4} = - \frac{2015}{4}$ ,

and multiplying both sides by $4$ :

$(2 k - 3)(2 n + 2017) = - 2015$ .

Because $2015 = (5^1) (13^1) (31^1)$ it has $(1+1)*(1+1)*(1+1) = 8$ factors, so $-2015$ has $16$ factors among the negative and positive integers. The expressions $2 k - 3$ and $2 n + 2017$ must both correspond to a factor of $-2015$ , since both are integers. Thus there are $16$ possible integral values for $2n+2017$ . $n$ must take the form $\frac{-2017 \pm m}{2}$ , where $m$ is one of the $8$ factors of $2015$ . So there are $16$ corresponding possible values for $n$ (they are: $-2016, -1, -1210, -807, -1086, -931, -1041, -976, -1024, -993, -1015, -1002, -1011, -1006, -1009, -1008$ ). So the answer is $16$ .

Just use script "Diophantine y=(3x+2018)/(2x+2017) on integers" in WolframAlpha, and you get list of 16 pairs (x,y) of integers, Answer=16.

One line in Mathematica:

Reduce[3x+2018==n(2x+2017),{x,n},Integers]

Yeilds all 16 solutions

I first changed $f(x')= \frac{3x'+2018}{2x'+2017}$ into the equivalent fraction $g(x) = \frac{2015+3x}{2015+2x}$ where x' = x - 1 This just made the equation look simpler without changing the mathematics

Then I looked at $2015+3x \equiv 0 mod(2015 + 2x)$ considering that $A+B \equiv 0 mod(2015 + 2x)$ we can look at the remainder from 2015 and 3x separately.

This allows for an interesting change of perspective 2015 mod(2015 + 2x) can be written as -2x mod(2015 + 2x) , because as 2015+2x increases 2015 stays constant and the remainder is simply -2x.

Rewriting $2015+3x \equiv 0 mod(2015 + 2x)$ gives $-2x + 3x \equiv 0 mod(2015 + 2x)$ or $x \equiv 0 mod(2015 + 2x)$ This brings us to an interesting result as 2015 + 2x increases by 2, x is only increasing by one, and therefore the congruence does not hold for positive x values. Essentially as x approaches infinity we get a horizontal Asymptote at 3/2.

So we can think about negative x values. As x gets more negative in g(x) the numerator decreases faster to 0 than the denominator. Then when x = -806, the numerator is -403 and the denominator is 403. As x gets more and more negative we can reason that the denominator will get closer to 0, resulting in multiple integer solutions. After it passes 0, both the numerator and denominator will be negative, and just like the positive x solutions there will be a point where the numerator is moving faster than the denominator but not fast enough to illicit integer multiples of it.

From here I ran a small bit of java code to calculate the numbers
public static void congruence() { ArrayList<Integer> solutions = new ArrayList<>(); int x = 0; while(2015-2 x >-10000) { for(int i =-10000;i<10000;i++){ if((2015-2 x)*i==x){ solutions.add(x); } } x++; } System.out.print(solutions + "\n" + solutions.size() + "\n"); newbe(solutions); }

public static void newbe(ArrayList<Integer> solutions) { for(Integer e:solutions){ int numerator = 3 (-e-1)+2018; int denominator = 2 (-e-1) + 2017; System.out.println(numerator + " " + denominator); } } }

The last x value I got was -2015 making $g(x) = \frac{-4030}{-2015} = \frac{2}{1}$ since g(x) is approaching a horizontal Asymptote of 3/2, there will not be a 1/1 fraction, and the 16 values the code generated are the only 16.

Another brute-force solution, using SAS this time (using more extreme values of x does not change the number of solutions):

data a; do x=-10000 to 10000; y= (3 x+2018) / (2 x+2017); if y-int(y)=0 then output; end; run;

proc contents data=a; run;

Let N be the result of (3x+2018)/(2x+2017).

Solving for x gives:

x=(2018-2017N)/(2N-3)

This converges to -1008.5 as N increases as -2017N/2N dominates.

Brute force inspection of positive and negative intergers for N as x approaches -1008.5 from above and below gives the result. Once we get x > -1009 and < -1008 from below and above we can stop. Not very sophisticated, but it gets there.