Is there integer :$n>0$ for which $\displaystyle\frac{n^4+(n+1)^4}{n²+(n+1)^2}$ is integer?

Question

I have tried to show if $$\displaystyle \gcd (n^4+(n+1)^4,n^2+(n+1)^2 )=1$$ for every positive integer $n$ using standard theorem in number theory as Bèzout and Gauss theorem but I don't succeed. I'm now interesting to seek for a fixed integer $n>0$ for which the ratio: $$\displaystyle\frac{n^4+(n+1)^4}{n^2+(n+1)^2}$$

is integer if it is possible ?

Thank you for any help.

Answer 1

$$ \frac{n^4+(n+1)^4}{n²+(n+1)^2} = n^2 + n + \frac32 - \frac1{2 (2 n^2 + 2 n + 1)} $$ This reduces the question to when is $\displaystyle\frac12 - \frac1{2 (2 n^2 + 2 n + 1)}$ an integer?

The only solutions are $n=0$ and $n=-1$.

Answer 2

Let $d(n)$ denote the gcd in question. Since both terms are odd, $d(n)$ is odd, Note that $$\left(n^2+(n+1)^2\right)\times \left(n^2-(n+1)^2\right)=\left(n^4-(n+1)^4\right)$$ Thus $$d(n)\,|\,2n^4\quad\&\quad d(n)\,|\,2(n+1)^4$$ Since $d(n)$ is odd, we deduce $$d(n)\,|\,n^4\quad\&\quad d(n)\,|\,(n+1)^4$$Clearly, though, $\gcd(n^4,(n+1)^4)=1$ so $d(n)=1$.

Answer 3

Hint $\ $ Specialize $\ a,\,b = n^2,\, (n\!+\!1)^2\ $ in the below Euclidean gcd algorithm computation

$\quad (a\!+\!b,a^2\!+\!b^2) \color{#c00}{\overset{\rm E}=} (a\!+\!b,\color{#90a}{2a^2}) = (a\!+\!b,2)\, $ when $\,(a,b) = 1,\ $ by $\,(a,a\!+\!b)\color{#c00}{\overset{\rm E}=} (a,b)=1$

Remark $ $ We twice used $\rm\color{#c00}{E} =$ Euclid's $\ (A, B) = (A, B')\ $ if $\ B\equiv B'\pmod{\!A}$

e.g. in the first case: $\ {\rm mod}\ a\!+\!b\!:\,\ \color{#0a0}{b\equiv -a}\,\Rightarrow\, a^2+\color{#0a0}b^2\equiv a^2+\color{#0a0}a^2\equiv \color{#90a}{2a^2}$