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Question: If $n$ is a positive number I write $a(n) =1+n+n^2$. Let $s$ be a fixed positive number. Is it true the $\gcd(a_n+1, a_{n+s})$ will be a divisor of $1+5s^2+s^4$?

For example if $n=1$ and $s=3$ then $\gcd(4,21)=1$ which is a divisor of $1+5*3^2+3^4=127$, since $1 |127$. On the other hand if $n=106$ then $\gcd(10922,11557)=127$ and $127 |127$.

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    Just computing stuff. Not sure where to start. I know that $a(n+s)=1+n+s+n^2+2ns+s^2$. – 2017-01-18
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    I would try to find a recurrence there – 2017-01-18

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$g = \gcd(a_n + 1, a_{n+s}) \\= \gcd(n^2 + n + 2, (n+s)^2 + (n+s) + 1) \\= \gcd(n^2 + n + 2, s^2 + 2ns + s - 1)$

So if we have $u \cdot (n^2 + n + 2) + v \cdot (s^2 + 2ns + s - 1) = g$, we can apply the Extended Euclidean Algorithm for polynomials to get:

$(\frac{4s^2}{s^4+5s^2+1}) \cdot (n^2 + n + 2) + (\frac{s^2-2ns-s-1}{s^4+5s^2+1}) \cdot (s^2 + 2ns + s - 1) = 1$

Multiply to flatten the expression:

$(4s^2) \cdot (n^2 + n + 2) + (s^2-2ns-s-1) \cdot (s^2 + 2ns + s - 1) = s^4+5s^2+1$

This means $\gcd(a_n + 1, a_{n+s})$ must divide $s^4+5s^2+1$, so the answer to your question is yes.

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    Actually you might have glossed over way to much for me to follow this answer directly. – 2017-01-18
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    The main steps I glossed over are contained in the Wikipedia link, which explains how to find $u$ and $v$. – 2017-01-18
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    Actually I am stuck here: $\gcd(n^2+n+2,(n+s)^2+(n+s)+1)=gcd(n^2+n+2,s^2+2ns+s-1)$. I am not sure how you eliminated terms? – 2017-01-18
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    $\gcd(a, b) = \gcd(a-b, b) = \gcd(a, b-a)$. In this case I subtracted the left side from the right. – 2017-01-18
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    This is what I have been looking for ! – 2017-01-18