Find the values of $n$ that make the fraction $\frac{2n^{7}+1}{3n^{3}+2}$ reducible.

Question :

Find 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 $k≠1$

If any one have any idea or hint please give me to see

Thanks!

Solution 1:

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 $435$.

Solution 2:

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 $Bn^C+D.$