Is there a combination of $a,b,c\in\mathbb{N}$ such that $\forall n\in\mathbb{N}\; \gcd(n^a+b,(n+c)^a+b)=1$?

Question

There is a related question on math.stackexchange, but this question here is different.

Let $n$ be a natural number ($n\in\mathbb{N}$). Calculate the value of this expression:

$$ m=\mathrm{gcd}(n^{17}+9,(n+1)^{17}+9) $$

Here $\mathrm{gcd}⁡(a,b)$ is the greatest common divisor of $a$ and $b$ with $a=n^{17}+9$ and $b=(n+1)^{17}+9$. The result of this expression is $m=1$ if $a$ and $b$ are coprime, and this is the case for any $n<n_0$ with

$$ n_0=8424432925592889329288197322308900672459420460792433 $$

which is a number with 52 decimal digits. If the value of $n$ is exactly this number, then the value of

$$ m=8936582237915716659950962253358945635793453256935559 $$

which is another number with 52 decimal digits. And of course you get the same result for any $n=n_0+k\cdot m$ where $k$ is any non-negative integer, which means that there are infinite many solutions to

$$ \mathrm{gcd}(n^{17}+9,(n+1)^{17}+9)\neq 1 $$

I did not study mathematics (this is why I have troubles understanding the answer to the linked question from above), but I am teaching computer science, and I used this expression to write this small Python program:

import math
# Initialization
n = 1
gcd = 1
# loop as long as gcd equals 1
while gcd == 1:
    n += 1
    gcd = math.gcd(n ** 17 + 9, (n + 1) ** 17 + 9)
# after the loop
print("n   =",n)
print("gcd =",gcd)
print("halting now")

I ask my students if this program will halt, and I ask them for suggestions of methods of how to answer this question. This turns out to be a good starting point to discuss Turing’s Halting problem. This program is well suited because it halts. But without knowing the correct answer in advance or understanding the math behind the problem, it is almost impossible to find the correct answer in a reasonable amount of time. (I can show that this program halts when I replace the first 1 with a value slightly below $n_0$.)

But last year, after a few minutes my students agreed on this answer: "Yes it will halt, because if it would run forever, you wouldn't have shown it to us." This is clever, but it takes away the motivation to suggest methods of figuring out whether or not it will halt.

So I want to give them a second program that is as similar as possible to the program shown here, but this second program should run forever. And there should be a proof that it can never halt. Then I would tell my students that exactly one of the two programs will halt, and I would ask them for methods to find out which of the two is the halting program.

And this brings me to my question:

Is there a combination of $a,b,c\in\mathbb{N}$ such that

$$ \forall n\in\mathbb{N}\left(\mathrm{gcd}(n^a+b,(n+c)^a+b)=1\right) $$ ?
An answer is preferred where $a,b,c$ are close to $17,9,1$. (Especially $c=1$ would be very fine.)

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Addendum (reaction to a comment):
There are infinite many trivial solutions: $a=1; b\in\mathbb{N}; c=1$. But this is too simple. My students would quickly find out, that $\mathrm{gcd}(n^1+b,(n+1)^1+b)=\mathrm{gcd}(n+b,n+1+b)=1$ for all $n$ and for all $b$, and that a program with these trivial parameters would never halt.

Answer 1

Certainly $b=0$ and $c=1$ (one of OP's comments says that "$b=0$ is fine") is enough to guarantee this, regardless of the value of $a$: $$\gcd(n^a,(n+1)^a)=1$$ holds because any prime $p$ that divides $n^a$ also divides $n$, and hence $(n+1)^a \equiv 1$ (mod $p$). So $n^a$ and $(n+1)^a$ can't have any common prime factor.

On the other hand, based on the reasoning in the linked question, less trivial examples with $b\neq 0$ and $c=1$ can only be found when the resultant of $n^a+b$ and $(n+1)^a+b$ is equal to $\pm 1$ for some positive integer $b$. This doesn't happen for $a=2,3,4$; but it does seem worth checking for higher values of $a$, and that check could be automated.