Is the greatest common divisor of n^17+9 and (n+1)^17 + 9, always = 1 for any integer value of n?

Good question. It sure looks like it.

I've crunched a lot of numbers looking for a value of n where it's not, but haven't found any. I used Wolfram Alpha up to n=20, and GCD is always 1.

The Polynomial GCD of n^17+9 and (n+1)^17 + 9 is 1, but I'm not sure how that would help.

I'd sure like to see a rigorous answer to this one.

It seems that n^p+9 and (n+1)^p + 9 are always relatively prime. I'm not sure what the role of 9 is.

It's not true that n^p + k and (n+1)^p + k are relatively prime for any prime p and natural number k, since GCD (6^13+ 1, 7^13 + 1) = 53. That tells you that any theorem about this, if there is any, can't be TOO general. (But change the 1 to anything greater, and I can't find a counterexample for exponent 13.)

It appears that there is always some n for which n^p+1 and (n+1)^p + 1 will have a GCD of ep+1, where e is some even number.

7 = GCD(5^3+9, 6^3+9)

11 = GCD(6^5+9, 7^5+9)

23 = GCD(10^11+9, 11^11+9)

53 = GCD(6^13+9, 7^13+9)

Maybe that's because n+1 and (n+1) + 1 are factors of n^p + 1 and (n+1)^p + 1, but why that would help isn't clear.

It's a trick question. The gcd is >1 for n = 8424432925592889329288197322308900672459420460792433. That's the lowest value of n (51 digits!).