SOLUTION: Find all integers n for which \frac{n^2 + n + 1}{n

SOLUTION: Find all integers n for which \frac{n^2 + n + 1}{n - 2 + n^3} is an integer.

Question 1209684: Find all integers n for which \frac{n^2 + n + 1}{n - 2 + n^3} is an integer.

Found 2 solutions by CPhill, math_tutor2020:
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let the given expression be equal to an integer $k$.
\[ \frac{n^2 + n + 1}{n^3 + n - 2} = k \]
\[ n^2 + n + 1 = k(n^3 + n - 2) \]
\[ n^2 + n + 1 = kn^3 + kn - 2k \]
\[ kn^3 - n^2 + (k-1)n - (1+2k) = 0 \]
If $k=0$, then $n^2 + n + 1 = 0$, which has no integer solutions.
If $n=0$, then $\frac{1}{-2} = k$, which is not an integer.
If $n=1$, then $\frac{3}{0}$, which is undefined.
If $n=2$, then $\frac{7}{8}$, which is not an integer.
If $n=3$, then $\frac{13}{28}$, which is not an integer.
If $n$ is large, then $n^3$ dominates, so $k$ must be close to 0.
If $k=1$, we have $n^3 - n^2 + 0n - 3 = 0$.
$n^3 - n^2 - 3 = 0$.
If $n=2$, $8-4-3 = 1 \ne 0$.
If $n=3$, $27-9-3 = 15 \ne 0$.
If $n^3 + n - 2 = 1$, then $n^3 + n - 3 = 0$.
If $n=1$, $1+1-3 = -1$.
If $n=2$, $8+2-3 = 7$.
If $n^3 + n - 2 = -1$, then $n^3 + n - 1 = 0$.
If $n=0$, $-1$.
If $n=1$, $1$.
If $n=2$, $n^3+n-2 = 8+2-2 = 8$. Then $\frac{4+2+1}{8} = \frac{7}{8}$.
Consider the case where $n$ is large. Then $k \approx \frac{n^2}{n^3} = \frac{1}{n}$.
If $n$ is large, $k$ must be close to 0.
If $k=0$, $n^2+n+1=0$, no integer solution.
If $n=1$, $\frac{3}{0}$ undefined.
If $n=2$, $\frac{7}{8}$.
If $n=3$, $\frac{13}{28}$.
If $n=-1$, $\frac{1-1+1}{-1-1-2} = \frac{1}{-4}$.
If $n=-2$, $\frac{4-2+1}{-8-2-2} = \frac{3}{-12} = -\frac{1}{4}$.
If $n=-3$, $\frac{9-3+1}{-27-3-2} = \frac{7}{-32}$.
Final Answer: The final answer is $\boxed{no solutions}$

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Replace each instance of n with x.

The problem is
Find all integers x for which (x^2+x+1)/(x^3+x-2) is an integer

Consider these two functions
f(x) = abs(x^2+x+1)
g(x) = abs(x^3+x-2)
where "abs" is shorthand for "absolute value".
Vertical bars can be used in place of abs notation.

The goal is to see when f/g is an integer for integer values of x.

Use a graphing tool of your choice, such as Desmos, to get this diagram

f(x) in green
g(x) in blue
Note that all of y = x^2+x+1 is entirely above the x axis, so abs(x^2+x+1) is identical to x^2+x+1. The absolute value bars have no effect on f(x).
The abs is much more useful with g(x) since some parts of x^3+x-2 go below the x axis.

The two functions intersect at these approximate locations
(0.3926, 1.5468)
(1.8637, 6.3371)
Which shows that the solutions to f(x) = g(x) are roughly x = 0.3926 and x = 1.8637

According to the graph, if 0.3926 < x < 1.8637 then the green curve is above the blue curve.
This is when f(x) > g(x).
Otherwise, f(x) < g(x)

So we only need to search for integers in the interval 0.3926 < x < 1.8637
This narrow interval contains the integer x = 1 and no other integers.

f(x) = x^2+x+1
f(1) = 1^2+1+1
f(1) = 3
g(x) = x^3+x-2
g(1) = 1^3+1-2
g(1) = 0
h(x) = ( f(x) )/( g(x) )
h(1) = ( f(1) )/( g(1) )
h(1) = 3/0
h(1) = undefined
Recall we cannot divide by zero.

So f/g is not an integer, and not even defined, when x = 1.

If x is an integer other than 1, then f/g will not be an integer.
Why not? Because the denominator g(x) is larger in absolute value compared to f(x).
As an example, f/g = 7/8 when x = 2
Another example is f/g = 3/(-12) = -1/4 when x = -2

So f/g is never an integer when x is an integer.

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