TITLE: (Non?)-uniqueness of sums of squares
QUESTION [11 upvotes]: (I've had almost no exposure to number theory, so please keep answers as elementary as possible.)
Write $\mathbb{N} = \{0,1,2,3,\ldots\}$ for the natural numbers. Then every element of $\mathbb{N}$ can be expressed as a sum of squares. For example:
$$6 = 1+1+1+1+1+1$$
Usually, we can get away with fewer terms in our sum. For instance, in the preceeding example, we can get down from 6 terms to 3 by writing $6 = 4+1+1.$ So by the potency of $n \in \mathbb{N}$, let us mean the least $k \in \mathbb{N}$ such that $n$ can be expressed as a sum of $k$-many squares. Hence the potency of $6$ is $3.$

Question. Let $n \in \mathbb{N}$ denote a natural number with potency $k$. Suppose $x$ and $y$ are sequences of length $k$ in
  $\mathbb{N}$ such that $$n=\sum_{j = 1}^k x_j^2 = \sum_{j = 1}^k y_j^2.$$
Are the sequences $x$ and $y$ necessarily equal up to a reordering of their terms?

I'm guesssing not, but haven't been able to find a counterexample.
Further information.

If $n$ is a prime of potency $2$, then the sequences $x$ and $y$ are equal up to reordering; see here.
Here we learn that the corresponding statement for higher powers is false, since $1^3+5^3+5^3=2^3+3^3+6^3.$

REPLY [3 votes]: To expand on some of the other comments:
Any number can be written as a sum of $4$ squares. In fact, the number of ways to write the number $n$ as a sum of $4$ squares (more precisely, the number of tuples $(x,y,z,w)\in \mathbb{Z}^4$ with $x^2 + y^2 + z^2 + w^2 = n$) is $$8\sum_{d\mid n,\, 4\not\mid d}d$$ (I don't know of a nice reference for this fact, nor a proof that does not use modular forms unfortunately).
Now note that these numbers tend to be much larger than they would be if all such ways were rearrangements of each other (the above sum tends to grow with $n$, and there are infinitely many numbers which are not the sum of $3$ squares).