Recreatii Matematice 2/2016, Problem VII.208

Problem:
Prove that the number $$N=2016^{n+1}-2015n-2016$$ has at least $27$ divisors for any $n \in \mathbb{N}^*$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
We have $$\begin{array}{lll} N&=&2016\cdot(2016^n-1)-2015n\\&=&2016\cdot2015\cdot(2016^{n-1}+2016^{n-2}+\ldots+1)-2015n\\&=&2015\cdot(2016^n+2016^{n-1}+\ldots+2016-n)\\&=&2015\cdot[(2016^n-1)+(2016^{n-1}-1)+\ldots+(2016-1)] \end{array}$$
Since $2016^n-1$ is divisible by $2015$ for all $n \geq 1$, we have that $2015^2 \ | \ N$. As $2015=5\cdot13\cdot31$, then $2015=5^2\cdot13^2\cdot31^2$, which has $27$ divisors. The conclusion follows.

Recreatii Matematice 1/2016, Problem VIII.201

Problem:
Prove that the number $2(n^4-n^2+1)$ is the sum of two perfect squares for all $n \in \mathbb{N}$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:We have $$\begin{array}{lll}2(n^4-n^2+1)&=&2n^2(n^2-1)+2\\&=&2n\cdot(n+1)n(n-1)+2\\&=&[(n+2)+(n-2)](n+1)n(n-1)+2\\&=&(n+2)(n+1)n(n-1)+1+(n+1)n(n-1)(n-2)+1 \\&=&[(n+2)(n-1)(n+1)n+1]+[(n+1)(n-2)n(n-1)+1]\\&=&[(n^2+n-2)(n^2+n)+1]+[(n^2-n-2)(n^2-n)+1]\\&=&(n^2+n-1)^2+(n^2-n-1)^2. \end{array}$$