How do you compute the $\gcd(1+n+n^2,1+n+n^2+s+2ns+s^2)$

Question

I would like to prove the following claim which I think is true:

Claim: Let $n,$ $m$ and $s$ be positive numbers. Fix $s$, then for every positive number $n$ the $\gcd(1+n+n^2,1+n+s+n^2+2ns+s^2)$ will be equal to a divisor of $1+5s^2+s^4.$

For example for every positive number $n$ if we set $s=8$ the $\gcd(1+n+n^2,73+n+16n+n^2)=1,7,631$ or $4417$. We can see that $1+5*8^2+8^4=4417$ and $4417=7*631$.

I came to the claim by moving numbers around in GAP. If it is wrong a counter example would be awesome.

Answer 1

Take $n=1$ and $s=3$. Then $$\displaylines{ 1+n+n^2=3\cr 1+n+n^2+s+2ns+s^2=21\cr \gcd(3,21)=3\cr 1+5s^2+s^4=127\cr 3\ \hbox{is not a factor of}\ 127\ .\cr}$$

Answer 2

No way a statement like this can be right.

$$\gcd(1+n+n^2, 1+n+s+n^2+2ns+s^2) = \gcd(1+n+n^2, s(1+2n+n^2)) = \gcd(1+n+n^2, sn) = \gcd(1+n+n^2, s)$$. So simply take $s = n^2+n+1$, then it can never be $s |1+5s^2+s^4$, unless $s=1$.