An elementary identical equation

Given positive integer $n$. Let perfect numbers be integers which can be written as $a_1^n+\cdots+a_n^n$, where $a_i\in\mathbb{Z}$. Prove that the product of $n$ perfect numbers is still perfect number.

I think the proof is to directly construct $a_i$ of the product, but I can't remember the detail.

Thanks for your clarification (Brahmagupta-Fibonacci identity), which gives me a way to answer. This is readily disproven for $n=3$ thanks to the sum of three cubes problem:

For $a, b, c \in \mathbb{Z}$, we know that $a^3 + b^3 + c^3 \not \equiv 4$ or $5 \pmod 9$. However, if we have $x, y, z \in \mathbb{Z}$, $x \equiv 1 \pmod 9, y \equiv 2 \pmod 9,$ and $z \equiv 2 \pmod 9$, then $x,y,$ and $z$ can all be expressed as the sum of three cubes, but their product, $xyz \equiv 4 \pmod 9$, cannot. QED.

Similarly, all integer $x^4 \equiv 0,1 \pmod{16}$, so all sums of four fourth powers will be in $\{0,1,2,3,4\} \pmod{16}$. But $1 \cdot 2 \cdot 2 \cdot 3 \equiv 12 \pmod{16}$, which cannot be a sum of four fourth powers. I suspect one can find similar analogous arguments for other powers.