How can we prove that $\gcd((n^4) + (n+1)^4 , (n+1)^4 + (n+2)^4) = 1$?

Question

I have found through experimentation that the consecutive sums of consecutive natural numbers raised to certain powers$(1,2,4)$ are always coprime. I was looking for modular multiplicative inverses and came across this facet. If you define a sequence $a(n): n^k + (n+1)^k$ , then $a(n)$ and $a(n+1)$ are coprime when $k=1,2$ or $4$ but no others. How can we prove this?

Since $a(n)$ and $a(n+1)$ are coprime when $k=1,2$ or $4$, their modular multiplicative inverse exists, and are sequences in the OEIS already.

I sincerely hope that this is moderately worthy and not cringe.

Answer 1

Suppose $d$ is a common factor of $n^4+(n+1)^4$ and $(n+1)^4+(n+2)^4$. Then $$d\bigg|(n+2)^4-n^4\\ \implies d\bigg|4(n+1)(n^2+(n+2)^2)$$ But note that $4(n+1)$ and $d$ are coprime (since both of the given numbers are odd, hence $d$ is also odd, also, if $k$ is a common factor of $d$ and $n+1$, then $k$ also divides $n^4$ which is a contradiction, since, $k$ and $n$ are coprime). Hence, $$d\bigg|n^2+(n+2)^2\tag1\\ \implies d\bigg|n^4+2(n+1)^4+(n+2)^4-\Big(n^2+(n+2)^2\Big)^2\\ \implies d\bigg|2(n+1)^4-2n^2(n+2)^2\\ \implies d\bigg|\Big((n+1)^2-n(n+2)\Big)\Big((n+1)^2+n(n+2)\Big)$$ $$\implies d\bigg|2n^2+4n+1\tag2$$ Subtracting the second equation from the first one, we get, $$d\bigg|3\tag3$$ Hence, $d$ is either $1$ or $3$. Since exactly one of $n, n+1$ and $n+2$ is divisible by $3$, we are done.

EDIT

As N.S. pointed out in comments, after $(1)$, if one observes that $$\Big(n^2+1\Big)\Big(n^2+(n+2)^2\Big)-\Big(n^4+(n+1)^4\Big)=3$$ then $d$ being a common factor of both terms in LHS, is also a factor of RHS and we again reach $(3)$.

Answer 2

Given $a(n)=(n+1)^4+n^4$.

You can check that: $$(4 n^3 + 20 n^2 + 39 n + 29)\cdot a(n) - (4 n^3 + 4 n^2 + 7 n + 1)\cdot a(n+1) = 12$$

Therefore any common factor of $a(n)$ and $a(n+1)$ must divide $12$.

You can find this by using the extended euclidean gcd algorithm applied to polynomials. I confess I did not do that by hand, but used Wolfram Alpha.

So now you only need to prove that $2$ and $3$ can never be a common factor. It is easy to see that $a(n)$ is odd whether $n$ is odd or even, and also that it is never a multiple of 3 regardless of whether $n\equiv 0,1,2\mod {3}$.

Similarly if $a(n)=(n+1)^2+n^2$ then you have

$$(n+2) \cdot a(n) - n\cdot a(n+1) = 2$$

so the only possible common factor could be 2, but this can be ruled out because $a(n)$ is always odd.

Answer 3

A quick mental way: suppose prime $p\mid f_n=(n\!-\!1)^4+n^4$ and $\,p\mid f_{n+1}.\,$ Then

$\!\bmod p\!:\ (n\!-\!1)^4 \equiv -n^4 \equiv (n\!+\!1)^4$ $\Rightarrow\, 0 \equiv (n\!+\!1)^4-(n\!-\!1)^4 \equiv 8n(n^2\!+\!1)$

contra $\,p\nmid 2\,$ by $\,f_n\equiv \color{#c00}{\bf 1}\pmod{\!2},\ $ $ p\nmid n\,$ by $\,f_n\equiv \color{#c00}{\bf 1}\pmod{\!n},\,$ and $\,p\nmid n^2\!+\!1\,$ else $\,p^{\phantom{|^{|^|}}}\!\!\mid\, f_n\bmod n^2\!+\!1 = -3,\,$ so $\,p=3\mid n^2\!+\!1,\,$ contradiction.