We know that :
"two polynomials in $\mathbb{F}[x]$ have a common root (possibly in some field extension of $\mathbb{F}$)" $\Leftrightarrow$ "the resultant of these two polynomials is zero"
which implies that for some prime $p$:
"there is a positive integer $n$ such that $\gcd\big(f(n), g(n)\big)=p$" $\Rightarrow$ "the polynomials $f,g$ have a common root $\mod p$" $\Rightarrow$ "their resultant is zero $\mod p$" $\Leftrightarrow$ "their resultant is a multiple of $p$"
(Here, we view the polynomials $f(x)=x^5+3$, $g(x)=(x+1)^5+3$ as polynomials of $\mathbb{Z}_p[x]$)
Thus, it suffices to investigate the prime divisors of the resultant: if $r$ is a divisor of the resultant $Res\big(n^a+b,(n+1)^a+b\big)=k$, then any solution of the simultaneous polynomial equations:
$$n^a+b \equiv 0 \mod r \\
(n+1)^a+b \equiv 0 \mod r $$
provides a value of $x=n$ for which: $\gcd\big(n^a+b,(n+1)^a+b\big)\geq r>1$.
Let us apply the above to the current problem. Utilizing Mathematica we get:
In[1]:= Resultant[x^5 + 3, (x + 1)^5 + 3, x]
Out[1]= 258751
and since the prime factorization of $ \ 258751=41*6311$, we will solve the system for all divisors of the resultant i.e. for $41, 6311, 258751$
The solutions can be found with standard methods, so we get
$$
\bigg\{
\begin{array}{c}
n^5 + 3\equiv 0\mod 41 \\
% \\
(n + 1)^5 + 3\equiv 0\mod 41
\end{array}\bigg\} \Rightarrow n\equiv29\mod 41
$$
$$
\bigg\{
\begin{array}{c}
n^5 + 3\equiv 0\mod 6311 \\
% \\
(n + 1)^5 + 3\equiv 0\mod 6311
\end{array}\bigg\} \Rightarrow n\equiv3810\mod 6311
$$
Thus:
$$
\begin{array}{c}
\bigg\{
\begin{array}{c}
n^5 + 3\equiv 0\mod 258751 \\
% \\
(n + 1)^5 + 3\equiv 0\mod 258751
\end{array}\bigg\} \Leftrightarrow
\bigg\{
\begin{array}{c}
x\equiv 29 \mod 41 \\
% \\
x\equiv 3810 \mod 6311
\end{array}\bigg\}\Leftrightarrow \\
\\
\Leftrightarrow n\equiv22743\mod 258751
\end{array}
$$
since, by the Chinese Remainder theorem
$$
\begin{array}{c}
22743=\mathbf{41}*554+\mathbf{29} \\
% \\
22743=\mathbf{6311}*3+\mathbf{3810}
\end{array}
$$
Thus the integer values $n$ for which $n^5+3$, $(n+1)^5+3$ are not relatively primes, are given by the following sequences:
$$
\begin{array}{c}
n_{29,41}=29 + 41j \ \ (\textrm{ giving gcd }=41) \\
\\
n_{3810,6311}=3810 + 6311k \ \ (\textrm{ giving gcd }=6311) \\
\end{array}
$$
for all positive integers $j, k$, and their subsequence
$$
n_{22743,258751}=22743 + 258751i \ \ (\textrm{ giving gcd } = 258751=41*6311)
$$
for all positive integers $i$.
