Prove that for all positive integers $n$, $${n+3 \choose 2} +6{n+4 \choose 4} + 90{n+5 \choose 6}$$ is the sum of two perfect cubes.
Proposed by Alessandro Ventullo, Milan, Italy
Solution:
Let $$N=\dfrac{(n+5)!}{8(n-1)!}+\dfrac{(n+4)!}{4n!}+\dfrac{(n+3)!}{2(n+1)!}.$$
We have $$N=\dfrac{1}{8}[n(n+1)(n+2)(n+3)(n+4)(n+5)+2(n+1)(n+2)(n+3)(n+4)+4(n+2)(n+3)]$$
Set $x=n^2+5n$. Hence, $$\begin{array}{lll} N&=&\dfrac{1}{8}[x(x+4)(x+6)+2(x+4)(x+6)+4(x+6)]\\&=&\dfrac{1}{8}[x(x+4)(x+6)+2(x+4)(x+6)+4(x+4)+8]\\&=&\dfrac{1}{8}(x+4)(x(x+6)+2(x+6)+4)+1\\&=&\dfrac{1}{8}(x+4)^3+1\\&=&\left(\dfrac{n^2+5n+4}{2}\right)^3+1\\&=&\left(\dfrac{(n+1)(n+4)}{2}\right)^3+1,\end{array}$$ which is the sum of two perfect cubes since $(n+1)(n+4)$ is even for all $n \in \mathbb{N}^*$.
We have $$N=\dfrac{1}{8}[n(n+1)(n+2)(n+3)(n+4)(n+5)+2(n+1)(n+2)(n+3)(n+4)+4(n+2)(n+3)]$$
Set $x=n^2+5n$. Hence, $$\begin{array}{lll} N&=&\dfrac{1}{8}[x(x+4)(x+6)+2(x+4)(x+6)+4(x+6)]\\&=&\dfrac{1}{8}[x(x+4)(x+6)+2(x+4)(x+6)+4(x+4)+8]\\&=&\dfrac{1}{8}(x+4)(x(x+6)+2(x+6)+4)+1\\&=&\dfrac{1}{8}(x+4)^3+1\\&=&\left(\dfrac{n^2+5n+4}{2}\right)^3+1\\&=&\left(\dfrac{(n+1)(n+4)}{2}\right)^3+1,\end{array}$$ which is the sum of two perfect cubes since $(n+1)(n+4)$ is even for all $n \in \mathbb{N}^*$.
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