What is the largest integer with only one representation as a sum of five nonzero squares? It seems to be very well known that $33$ is the largest integer with zero representations as a sum of five nonzero squares. So it seems reasonable to me that as we go higher and higher, numbers have more and more representations as sums of five nonzero squares, and maybe there is a threshold above which all numbers have at least two such representations.
My question is, what is the largest integer with only one representation as a sum of five nonzero squares?
 A: It's getting late so I'll post an incomplete answer showing that an integer $\ge70$ can be represented as a sum of five squares in at least two different ways unless $n\equiv1\pmod8$.
The main tools:


*

*Three square theorem.

*An integer $\equiv3\pmod8$ can be represented as a sum of three odd squares (all of which are then non-zero). This is due to Gauss (equivalent to his result that every natural number can be written as a sum of three triangular numbers), and follows from the three square theorem, because the only way for three squares to add up to something $\equiv3\pmod 8$ is for all three squares to be odd. This is because $0,1,4$ are the only quadratic residues modulo $8$

*Similarly we see that any integer $\equiv6\pmod8$ can be written as a sum of three squares such that the squares in question can only be congruent to $4,1,1$ modulo $8$ respectively. In particular all those must be non-zero.


Let's roll. So I must assume that $n\equiv k\pmod8$ where $k=0,2,3,4,5,6,7$.
$\mathbf{k=0}$: By the second bullet both $n-4-1$ and $n-36-1$ are sums of three odd squares, so assuming $n\ge40$ we are done with this case.
$\mathbf{k=2}$: By the last bullet $n-16-36$ and $n-4-64$ are both sums of a single non-zero even square and two odd squares. Provided that $n\ge70$ we get two representations of $n$ as a sum
of three even squares and two odd squares. Because $4,16,36,64$ all appear, the two representations are different. 
$\mathbf{k=3}$: All we need to do here is to apply the second bullet to $n-4-4$ and $n-16-16$, so we only need to assume $n\ge35$.
$\mathbf{k=4}$: All we need to do here is to apply the second bullet to $n-16-1$ and $n-64-1$, which we are able to do provided $n\ge 68$. Note that the two representations are distinct, because the sole even squares in them are different.
$\mathbf{k=5}$: If all the numbers $n-1-1$, $n-9-9$, $n-25-25$ are all positive, then by the second bullet they are all sums of three odd squares. Necessarily at least two of the resulting representations of $n$ as a sum of five squares are different, because $1,9,25$ cannot all appear twice among a set of five squares. This case is settled, if $n\ge55$.
$\mathbf{k=6}$: By the last bullet if $n\ge38$, then both $n-4-4$ and $n-16-16$ can be written as sums of an even square $\equiv4\pmod8$ and two odd squares. Again we get two different representations.
$\mathbf{k=7}$: Here we simply apply the second bullet to both $n-4-16$ and $n-36-16$.
Again the resulting representations as sums of the prescribed two even squares and some three odd squares are different. This case is thus done, if $n\ge59$.
But the case $k=1$ gives me a headache. Here I cannot overcome the possibility of some of the participating squares becoming zero. One can try and find suitable representation as four squares for both $(n-1)/4$ and $(n-9)/4$, but the number of cases blows up, and it looks like it might lead to further splitting. May be morning (or somebody else!) will be wiser :-(
A: You may find it interesting to have a glance at this image: http://oeis.org/A025429/graph (Number of partitions of n into 5 nonzero squares).
 -- see also http://oeis.org/A080673 = largest numbers with exactly n representations as sum of five positive squares. 
