When is $\small \gcd(2n^7\!+1,{3n^3\!+2})>1,$ i.e. when is $\frac{2n^{7}+1}{3n^{3}+2}$ reducible? How can I ind the values of $n\in \mathbb{N}$  that make the fraction $\frac{2n^{7}+1}{3n^{3}+2}$ reducible ?
I don't know any ideas or hints how I solve this question.
I think we must be writte $2n^{7}+1=k(3n^{3}+2)$ with $k≠1$
 A: I have a solution, but I'm sure there's a better way to do this.  The greatest common divisor $g$ of of $2n^7+1$ and $3n^3+2$ must also divide $$3(2n^7+1)-2n^4(3n^3+2)=3-4n^4$$ then $g$ must also divide $$4n(3n^3+2)-3(4n^4-3)=8n+9$$  Then $g$ must divide $$ 3n^4(8n+9)-8(3n^3+2)=27n^2-16$$
Continuing in this manner, we eventually find that $g$ must divide $1163$ which is prime.  So any solution satisfies $$3n^3+2\equiv0\pmod{1163}$$
The only solution to this is is $n\equiv435\pmod{1163}$ which I found with a python script, though I imagine there's a way to do it with a pencil.
It's easy to verify that also $2\cdot435^7+1\equiv0\pmod{1163}$, so the complete solution is $$n\equiv435\pmod{1163}.$$ 
EDIT
Daniel Wainfleet's answer shows the right way to find $435$.     
A: Suppose $p$ is prime.
If $p$ divides $2n^7+1$ & $3n^3+2$
then $p$ divides $2(2n^7+1)-(3n^3+2)=n^3(4n^4-3)$ 
then $p$ divides $4n^4-3$ ( See Footnote )
then $p$ divides $2(4n^4-3)+3(3n^3+2)= n^3(8n+9)$
then $p$ divides $8n+9$ (See Footnote)
then $p$ divides $9(3n^3+2)-2(8n+9)=n(27n^2-16)$
then $p$ divides $27n^2-16$ (See Footnote)
then $p$ divides $9(27n^2-16)+16(8n+9)=n(243n+128)$
then $p$ divides $243n+128$ (See Footnote)
then $p$ divides $9(243n+128)-128(8n+9)=1163$
then $p=1163$ because $p$ and $1163$ are both prime
then $8n+9\equiv 0 \mod 1163$ so $n\equiv 435 \mod 1163$
So $\gcd (2n^7+1, 3n^3+2)>1\implies n\equiv 435 \mod 1163.$ And we may  verify  that $n\equiv 435 \mod 1163\implies 2n^7+1\equiv 3n^3+2\equiv 0 \mod 1163\implies \gcd (2n^7+1,3n^3+2)>1.$
Footnote. Suppose $A,B,C, D$ are integers with $A>0$ and $C>0,$ and  $p$ divides $n^A(Bn^C+D).$ Since $p$ is prime, $p$ divides $n^A$ or $p$ divides $Bn^C+D.$ Now if prime $p$ divides $n^A$ then $p$ divides $n$ and hence $p$ divides $2n^7$, BUT if $p$ also divides $2n^7+1$ then the prime $p>1$ divides $(2n^7+1)-(2n^7)=1,$ which is absurd. So instead, $p$ must divide $Bn^C+D.$
A: Let $d = \gcd(2n^7+1, 3n^3+2)$. Then since $2n^7+1 \ | \ 2^3n^{21}+1$ and $3n^3 + 2 \ | \ 3^7n^{21}+2^7$, we must have $$d \ | \ 3^7(2^3n^{21}+1) - 2^3(3^7n^{21}+2^7) \quad\Rightarrow\quad d \ | \ 1163.$$
Since $1163$ is a prime, if the fraction is reducible, $1163 \ | \ 3n^3 + 2$. Since $1163 \equiv 2 \pmod 3$, $n^3 \equiv -2\cdot 3^{-1} \equiv 587 \pmod{1163}$ has one unique solution modulo $1163$.
Fermat's little theorem tells you that $n \equiv n^{1163} \equiv n^{2325} \equiv n^{3\times 775}\pmod{1163}$. So the answer should be $n \equiv 587^{775} \pmod{1163}$.
This is hard to solve by hand, so if you believe saulspatz's computation that $n \equiv 435 \pmod{1163}$, that is all the possible solutions.
A: This gcd is computable purely mechanically by a slight generalization of the Euclidean algorithm which allows us to scale by integers $\,c\,$ coprime to the gcd during the modular reduction step, i.e.
$$\bbox[8px,border:2px solid #c00]{(a,b)\, = \,(a,\,cb\bmod a)\ \ \ {\rm if}\ \ \ (a,c) = 1}\qquad\qquad  $$
which is true since $\,(a,c)= 1\,\Rightarrow\, (a,\,cb\bmod a) = (a,cb)  = (a,b)\ $ by Euclid. When computing the gcd of polynomials $\,f(n),g(n)$ with integer coef's, we can use such scalings to force the lead coef of the dividend to be divisible by the lead coef of the divisor, which enables the division to be performed with integer (vs. fraction) arithmetic. Let's do that in the example at hand (but you may find it helpful to first study  the simpler examples here and here).
$\!\begin{align}{\rm{This\,\ yields}\!: \ \ \ }(3n^3\!+2,\,2n^7\!+1) &\,=\, (3n^3\!+2,\,\color{#0a0}{8n+9})\ \ \ {\rm by}\ [\![1]\!]\ \text{ below,  $\,c = 3^2$}\\[.4em]
&\,=\, (\color{#90f}{-1163},\ \  8n+9)\ \ \ {\rm by}\, [\![2]\!]\ \text{ below,  $\,c = 8^3$}\end{align}$
$\!\!\!\begin{align}
&\bmod\:\! \color{#c00}{3n^3\!+2}^{\phantom{|^{|^|}}}\!\!\!\!\!:\,\  3^2(2n^7\!+1)\equiv 2n(\color{#c00}{3n^3})^2\!+9\,\equiv\, \color{#0a0}{8n+9},\, \ {\rm by}\ \ \color{#c00}{3n^3\equiv -2}\,\qquad [\![1]\!]\\[.4em]
&\bmod\:\! \color{#0a0}{8n+9}\!:\ \   8^3(3n^3\!+2)\equiv 3(\color{#0a0}{8n})^3\!+ 2(8^3) \equiv  \color{#90f}{-1163},\ \ {\rm by}\ \ \color{#0a0}{8n\,\equiv \,-9}\:\!\qquad[\![2]\!] \end{align}$
So the gcd $\!=\! (1163,8n\!+\!9)\! >\! 1\! \iff\!$ prime $p\! =\! 1163\mid 8n\!+\!9\!$ $\iff\! \bbox[5px,border:1px solid #0a0]{n\equiv \color{90f}{435}\pmod{\!p}}\,$ by
$\!\!\bmod 1163\!:\,\ n\equiv\dfrac{\!\!-9}8\!\equiv 3\left[\dfrac{-3}8\right] \equiv 3\left[\dfrac{1160}8\right]\equiv 3[145]\equiv 435.\ \ $ [See here for 5 more ways]
Remark $ $ Above $\,k \bmod a\,$ denotes some "simpler" $k'$ such that $\,k'\equiv k\pmod{\!a},\,$ not necessarily the least nonnegative such value. Here we are "simplifying" by reducing the degree in $\,n,\,$ i.e. essentially using the Euclidean algorithm for polynomials (the scaling by $\,c\,$ corresponds to a fraction-free form of the algorithm using the Nonmonic Division Algorithm).
