What is the largest integer with only one representation as a sum of five nonzero squares?
Let $A=7225$ and note that $$\tag1 \begin{align}A&= 85^2\\&=84^2+13^2\\&=84^2+12^2+5^2\\&=84^2+12^2+4^2+3^2,\end{align}$$ but also $$\tag2\begin{align} A&=77^2+36^2\\&=60^2+60^2+5^2\\&=60^2+60^2+3^2+4^2.\end{align}$$ Also let $B=49A$ so that we get corresponding representations to $(1)$ and $(2)$ with all $x^2$ replaced by $(7x)^2$. By Lagrange, we know that every natural number can be written as the sum of up to four nonzero squares. Thus if $n>A$, we apply this to $n-A$ and find that, depending on how many squares we need for $n-A$, we can pick a suitable representation from $(1)$ to obtain a representation of $n$ as sum of five nonzero squares. As we have alternate choices from $(2)$, we obtain a second such representation, unless the representation we found for $n-A$ used four nonzero squares and the only five-square representation for $n$ we find is of the form$$\tag3n=85^2+a^2+b^2+c^2+d^2.$$ If $n>B$, we can repeat the argument and either find two representations of $n$ as sum of five squares, or we find that $$\tag4n=595^2+e^2+f^2+g^2+h^2.$$ But then $(3)$ and $(4)$ are two distinct representations - unless they coincide, i.e. both are in fact $$\tag{3,4} n=85^2+595^2+i^2+j^2+k^2.$$ But then we make use of $1^2+7^2=5^2+5^2$ and find a second representation $$\tag5 n=425^2+425^2+i^2+j^2+k^2.$$ We conclude that all numbers $n>B=354025$ allow at least two representations as sum of five nonzero integers. To determine the smallest bound that can be used in place of $B$ is a finite task and left as exercise to the reader. (And the reader should find that the true threshold is much lower: We have two representations for all $n>60$)
Remark: Encouraged by the numerical finding that the actual threshold is $\ll B$, I made a little search for a better $A$ and found that using $$\begin{align}A:=625&=25^2\\&=7^2+24^2=15^2+20^2\\ &=9^2+12^2+20^2=15^2+12^2+16^2\\ &=9^2+12^2+12^2+16^2=10^2+10^2+13^2+16^2\end{align}$$ and $B:=4A$ together with the observation $25^2+50^2=10^2+55^2$ allows to show (by the very same method as above) that there are two representations for all $n>2500$. This signifiacntly lessens the "manual" work required for finding the true threshold.
Generalization: Now Let $k\ge 0$ and $$\begin{align}A=169 &=13^2\\&=5^2+12^2\\&=3^2+4^2+12^2\\&=1^2+2^2+8^2+10^2. \end{align}$$ Assume that $n>(30k+1)^2A$ has only $k$ representations as sum of five squares. Each of these representations gives us $30$ sums obtained by using one up to four of its summands. There is at least one $m$ with $1\le m\le30k+1$ such that none of these $30k$ sums evaluates to $m^2A$. But then we obtain a new representation of $n$ from applying Lagrange to $n-m^2A$, contradiction. We conclude that all $n>(390k+13)^2$ have at least $k+1$ representations as sum of five nonzero squares.
OK, I'll write this down as a new answer because it
- uses Jyrki Lahtonen's idea to use the Three square theorem instead of the four square tehorem
- uses Jyrki's method of distinguishing by the remainder $n\pmod 8$
- uses my idea of subtracting numbers that can be written as different sums of squares to eat up zeroes.
- most of all, it produces the exact threshold with virtually no additional work!
The "main trick" is as follows: Assume $$ A=a^2+b^2=c^2+d^2+e^2=f^2+g^2+h^2+i^2$$ with $a,\ldots ,i\in\mathbb N$. Then if $n>A$ and $n\not\equiv A\pmod 8$ and $n\not\equiv A-1\pmod 8$, we can write $n-A$ as sum of three squares by the three squares theorem. Depending on how many of these three squares are zero (not all of them because $n-A>0$), we have $$n-A=\begin{cases}u^2&\text{or}\\u^2+v^2&\text{or}\\u^2+v^2+w^2\end{cases}$$ with $u,v,w\in\mathbb N$ and hence obtain a representation of $n$ as sum of five nonzero squares: $$n=\begin{cases}f^2+g^2+h^2+i^2+u^2&\text{or}\\ c^2+d^2+e^2+u^2+v^2&\text{or}\\ a^2+b^2+u^2+v^2+w^2.\end{cases} $$
Theorem. Every integer greater than $60$ has at least two distinct representations as sum of five nonzero squares.
Proof. Let $n\in\mathbb N$ with $n>61$ (sic!) and write $n=8r+k$ with $0\le k\le 7$.
Consider first the case that $k\in\{0,1,2,3,6,7\}$. As $61\equiv 5\pmod 8$, the observation $$\tag1\begin{align} 61=6^2+5^2=6^2+4^2+3^2&=5^2+4^2+4^2+2^2\\ &=7^2+2^2+2^2+2^2 \end{align}$$ allows us to apply the main trick and obtain a representation $$\tag2 n=a^2+b^2+c^2+d^2+e^2$$ of $n$ as sum of five nonzero squares. In fact, if four of the summands in $(2)$ add up to $61$, we already obtain two distinct representations, so we assume that $61$ is formed by either two or three of the summands in $(2)$. As also $53\equiv 5\pmod 8$ and $$\tag353 = 7^2+2^2 = 6^2+4^2+1^2=6^2+3^2+2^2+2^2,$$ we obtain another representation of $n$. If this is a different representation, we are done. Otherwise we need to check how the summands from $(1)$ and $(3)$ can occur in $(2)$ and we distinguish cases depending on the summands these two sub-sums have in common:
- None: Then four or five of the summands in $(2)$ add up to $61+53=114$. Replacing these with $114=9^2+5^2+2^2+1^2=9^2+4^2+3^2+2^2+1^2$, we therefore obtain a distinct representation (it is distinct because the number of $9^2$'s occuring differs)
- $6^2$: Then four or five squares add up to $61+53-36=78$. Replacing these per $78=8^2+2^2+2^2 +1^2=6^2+6^2+2^2+2^2$ gives us a distinct representation (distinct because the number of $6^2$'s differs).
- $4^2$: A summand $4^2$ occurs in $(3)$ only in $6^2+4^2+1^2$ and in $(1)$ only in $6^2+4^2+3^2$ (remember that the case of four summands in $(1)$ has already been dealt with). Thus we we can refer to the case "$6^2+4^2$" below.
- $3^2$: Again, we conclude that $6^2$ is also a common summand, so see "$6^2+3^2$" below.
- $6^2+4^4$: We conclude that $$\tag4n=\rlap{\underbrace{3^2+6^2+4^2}_{61}}\hphantom{3^2+}\overbrace{\hphantom{6^2+4^2}+1^2}^{53}+a^2$$ for some $a\in\mathbb N$. Since $a^2\bmod 8$ is in $\{0,1,4\}$, we conclude that $k\in\{2,6,7\}$. If $a=1$, we have the two representations $n=63=6^2+4^2+3^2+1^2+1^2=5^2+4^2+3^2+3^2+2^2$. If $a>1$, then $n>65$ and as $65\equiv 1\pmod 8$ and $k\notin\{0,1\}$, we use the main trick with $$\tag5 65 = 8^2+1^2 = 6^2+5^2+2^2=6^2+4^2+3^2+2^2$$ to obtain a representation of $n$. This can conincide with the representation $(4)$ only as follows (with alternate representations exhibited): $$\begin{align}n&= 6^2+4^2+3^2+\overbrace{1^2+\underset{=a^2}{8^2}}^{65}=10^2+3^2+3^2+2^2+2^2\quad\text{or}\\ n&= 1^2+\overbrace{6^2+4^2+3^2+\underset{=a^2}{2^2}}^{65}=\hphantom{1}4^2+4^2+4^2+3^2+3^2.\end{align}$$
- $6^2+3^2$: This can happen only if $n=\rlap{\underbrace{4^2+6^2+3^2}_{61}}\hphantom{4^2+}\overbrace{\hphantom{6^2+3^2}+2^2+2^2}^{53}=69$, but then also $n=7^2+3^2+3^2+1^2+1^2$.
This completes the proof for the case $k\in\{0,1,2,3,6,7\}$. Assume now that $k\in\{ 4,5\}$. From $$\begin{align} 50&=5^2+5^2=5^2+4^2+3^2=4^2+4^2+3^2+3^2\\ &=7^2+1^2\hphantom{{}=5^2+4^2+3^2}=6^2+3^2+2^2+1^2\end{align}$$ with $50\equiv 2\pmod 8$ we see that the main trick can be applied and we obtain a representation of $n$ as sum of five nonzero squares. Unless what we obtain is $n=5^2+4^2+3^2+a^2+b^2$, we already get two distinct representations and are done. Doing the same with $$26 = 5^2+1^2 = 4^2+3^2+1^2=3^2+3^2+2^2+2^2$$ (where also $26\equiv 2\pmod 8$), we get another representation unless the following coincidence happens:
$$\tag6n=\rlap{\underbrace{4^2+3^2+5^2}_{50}}\hphantom{4^2+3^2+}\overbrace{\hphantom{5^2}+1^2}^{26}+a^2=\rlap{\underbrace{5^2+4^2+3^2}_{50}}\hphantom{5^2+}\overbrace{\hphantom{4^2+3^2}+1^2}^{26}+a^2. $$ If $a$ is even, we get $n\equiv 3\pmod 4$ contradicting $k\in\{4,5\}$; hence $a$ is odd. As $n>61$, we conclude $a\ge 5$, hence $n>74$ and from $$ 74 = 7^2+5^2=7^2+4^2+3^2=6^2+6^2+1^2+1^2$$ (with $74\equiv 2\pmod 8$ as well) we obtain another representation unless the following conincidence occurs: $$ n= 4^2+3^2+1^2+\overbrace{5^2+\underset{=a^2}{7^2}}^{74}=5^2+1^2+\overbrace{4^2+3^2\underset{=a^2}{7^2}}^{74}=100,$$ but then we have the second representation $$n = 9^2+4^2+1^2+1^2+1^2.$$
In summary, we have shown that for all $n>61$ there exist at least two distinct representations of $n$ as sum fo five nonzero squares. From $$ 61 = 7^2+3^2+1^2+1^2+1^2 = 5^2+5^2+3^2+1^2+1^2$$ we see that the result is in fact true for all $n>60$. $_\square$
On the other hand, one verifies that the representation $$\tag7 60=5^2+4^2+3^2+3^2+1^2$$ is unique, hence that the bound $60$ in the theorem is sharp, either by exhaustive search or as follows:
Assume $60=a^2+b^2+c^2+d^2+e^2$ with $a\ge b\ge c\ge d\ge e$. Then $4\le a\le 7$ because $5\cdot 3^2<60<8^2$. If $a=7$, then $60-7^2-3\ge b^2\ge \frac{60-7^2}{4}$ implies $b=2$, but $60-7^2-2^2=7$ cannot be written as sum of three squares. If $a=6$, then $60-6^2-3\ge b^2\ge \frac{60-6^2}{4}$ implies $b=3$ or $b=4$, but $60-6^2-4^2=8$ and $60-6^2-4^2=15$ cannot be written as sum of three nonzero squares. If $a=4$ then also $b=4$ because $4^2+4\cdot 3^2<60$ and $c=4$ because $2\cdot4^2+3\cdot 3^2<60$; but $60-3\cdot 4^2=12$ is not the sum of two squares. Remains $a=5$, $b^2+c^2+d^2+e^2=35$. From $4\cdot 3^2>35$ we conclude $e\le 2$. Then from $3\cdot 3^2+2^2<35$, we conclude $b\ge 4$. If $b=5$ then $10=c^2+d^2+e^2$ is written as sum of three nonzero squares, which is not possible. Therefore $b=4$, and we have to write $19$ as sum of three, necessarily odd, squares. The only solution leads to $(7)$.
Generalization. Let $r(n)$ denote the number of representations of $n$ as sum of five nonzero squares.
- If $n>5408$ then $r(n)\ge \lfloor\frac18\sqrt{n-101}\rfloor$; if additionally $n\not\equiv 1\pmod 8$ then $r(n)\ge\frac{n}{720}$.
- If $k\ge10$ and $n\ge 64k^2+101$ then $r(n)\ge k$.
- If $k\ge 7$, $n>720(k-1)$, and $n\not\equiv 1\pmod 8$, then $r(n)\ge k$.
Proof. As pointed out by Jyrki Lahtonen in his answer, unless $n\equiv 1\pmod 8$, we obtain a representation of $n=a^2+b^2+c^+d^2+e^2$ as sum of five nonzero squares for each choice $(a,b)$ with $a^2,b^2<\frac n2$ and $a\bmod 4$, $b\bmod 4$ have a prescribed value (depending on $n\bmod 8$). For $a$ (and also for $b$) there are at least $\left\lfloor\frac14\sqrt{\frac{n-1}2}\right\rfloor$ choices. Each representation is found repeatedly at most $5\cdot 4$ times (the ways $a$ and $b$ can be picked among the five summands). This gives us at least $$ \tag8r(n)\ge\left\lceil\frac1{20}\left\lfloor \frac14\sqrt{\frac{n-1}2}\right\rfloor^2\right\rceil\sim\frac{n}{640}$$ representations. To make the "$\sim$" more explicit, one verifies $$r(n)\ge\begin{cases}\frac n{784}&\text{if $n\ge1000$ (with $n=1568$ being critical)},\\ \frac n{720}&\text{if $n\ge3873$ (with $n=7200$ being critical)},\\ \frac n{648}&\text{if }n>10^6. \end{cases} $$
We now come to the case $n\equiv 1\pmod 8$. If $a\equiv 0\pmod 4$ and $a^2<n-100$, then $n-10^2-a^2>0$ and $\equiv 5\pmod 8$ can be written as sum of up to three nonzero squares. One of these squares must be $\equiv 1\pmod 8$, one must be $\equiv 4\pmod 8$, and the third square, if any, is $\equiv 0\pmod 8$. Hence either $$\tag{9a}n=8^2+6^2+a^2+b^2+c^2\text{ with $b\equiv 1\pmod 2, c\equiv 2\pmod 4$}$$ or $$\tag{9b}n=10^2+a^2+b^2+c^2+d^2\text{ with additionally $d\equiv 0\pmod 4$}.$$ Let us agree that we use $(9a)$ only if for the given choice of $a$ no representation of type $(9b)$ exists. There are at least $\left\lfloor \frac14\sqrt{n-101}\right\rfloor$ choices for $a$. Distinct $a$ can never lead to the same type $(9a)$ representation (only $a$ is a multiple of $4$); distinct $a$ can lead to the same representation of type $(9b)$ only if $a\leftrightarrow d$ are swapped; by our preference for $(9b)$ over $(9a)$, it is not possible for two choices of $a$ leading to the same representation where one is of type $(9a)$ and the other is of type $(9b)$. We conclude that $$ \tag{10}r(n)\ge\frac 12\left\lfloor \frac14\sqrt{n-101}\right\rfloor\ge \left\lfloor \frac18\sqrt{n-101}\right\rfloor.$$ It is clear that this bound is lower than the one found in $(8)$ if $n$ is large enough. One verifies (standard estinmates for floor/ceiling will work, but, hey, just look at the graphs of the functions) that the largest $n$ where the bound $(8)$ is lower than the bound $(10)$ is $n=5408$, hence the first claim. The other claims follow by solving the inequalities for $n$. $_\square$
Remark: Donovan Johnson (2013) reports on OEIS (sequence A080673) that no $n\le 10^6$ with $r(n)=188$ exists. According to the estimate in the generalization, extending the search range up to $2286245$ (where one may restrict to $n\equiv 1\pmod 8$) would prove That no $n$ with $r(n)=188$ exists at all (or find all such $n$). In the proof above I made use only of $10^2=6^2+8^2$. One may find more representations with using more even-sided pythagorean triangles, such as $20^2=12^2+16^2$ or $26^2=24^2+10^2$. While I wanted to avoid the more complex bookkeeping about duplicates, persuing this might show that D. Johnson's finding is sufficient to conclude that $r(n)\ne 188$ for all $n$.
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 :-(