, 2005

Solution:
If $n=1$, then the claim is true; the only solutions for $a, b, c, d$ are, apart from order, the quadruples $(1,1,1,5)$, $(1,3,3,3)$ and $(2,2,2,4)$. For every $n \geq 1$, $4^{n} \cdot 7$ is divisible by $4$. Since the square of a natural number can only have remainder $0$ or $1$ upon division by $4$, only numbers of the same residue class modulo $4$ are possible for $a, b, c, d$.
If $a, b, c, d$ are (all) odd, then the right-hand side is indeed divisible by $4$, but not by $8$, and certainly not by $16$. In this case, therefore, only $n=1$ is possible. If $n>1$, then the four numbers on the right-hand side must accordingly all be even.
Assuming the property that all square numbers exceed $4^{n-1}$ does not hold for all $n$, then there exists a smallest $k$ ($k \in \mathbb{N}$) for which it does not hold. Since it holds for $n=1$, $k$ must be greater than $1$. But then (see above) $a, b, c, d$ are all even. One could then divide both sides by $4$, and the property would also fail to hold for $k-1$, which, however, cannot be the case due to the minimality of $k$. The claim therefore holds for all natural numbers $n$, $n \geq 1$.