Prove for all integer $n > 1$ that if $n | 34$, then $n+5$ and $n^2+$9 are coprime

Question

My attempt was suppose $\gcd (n+5, n^2+9) = d$ where $d$ is an integer, hence $d|n+5$ and $d|n^2+9$ but I don't know how could I rearrange this to make it $1$

Answer 1

The easy way is following. Since $n\in\{1,2,17,34\}$ check the statement for each $n$:

$$n=1: \gcd(6,10)=2$$

$$n=2: \gcd(7,13)=1$$

$$n=17: \gcd(22,298)=2$$

$$n=34: \gcd(39,1165)=1$$

So the statement is true only for even $n$.

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"Hard way": Say exists prime $p\mid n+5$ and $p\mid n^2+9$, then $p\mid n^2-25$ so $$p\mid (n^2+9)-(n^2-25) = 34$$

This means $p= 2$ or $p=17$.

If $p=17$ we get $n+5\geq 17$ so $n=17$ or $n=34$ and thus $p\mid n$ so $p\mid 5$ which is not true.

Case $p=2$ is possible iff $n$ is odd.

Answer 2

[Note: this answer presumes that the question intended $\,34\mid n,$ not $\,n\mid 34$ as written.]

Hint:

If $d$ divides both $n+5, n^2+9$

$d$ must divide $n^2+9-(n+5)(n-5)=34$

So, $(n+5,n^2-9)$ must divide $34$

But $34\mid n\implies(n+5,34)=1$