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 $k≠1$

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 $Bn^C+D.$
