Find all pairs $(n, r), n$ integer $>0, r$ real, such that $(x+1)^{n}-r$ is divisible by $2 x^{2}+2 x+1$
$2 x^{2}+2 x+1$ admitting two distinct roots, namely two complex roots $\frac{-1+i}{2}$ and $\frac{-1-i}{2}$, it is necessary and sufficient that each of them is a root of $(x+1)^{n}-r$, so that $\left(\frac{1+i}{2}\right)^{n^{2}}=$ $r=\left(\frac{1-i}{2}\right)^{n}$. Now $\left(\frac{1+i}{2}\right)^{n}$ is real if and only if $n$ is a multiple of 4, in which case, if $n=4 k$, $\left(\frac{1+\mathrm{i}}{2}\right)^{4 \mathrm{k}}=\left(\frac{1-\mathrm{i}}{2}\right)^{4 \mathrm{k}}=\left(\frac{-1}{4}\right)^{\mathrm{k}}$. The solutions sought are therefore all the pairs: $\left(4 \mathrm{k},\left(\frac{-1}{4}\right)^{\mathrm{k}}\right)$