# Sum of one, two, and three squares

If a square $n^2$ can be written as the sum of two nonzero squares as well as the sum of three nonzero squares, then can we conclude that it can be written as the sum of any number of nonzero squares up to $n^2 - 14$ nonzero squares?

Example: $13^2 = 12^2 + 5^2 = 12^2 + 4^2 + 3^2$. But also $13^2 = 11^2 + 4^2 + 4^2 + 4^2$ and $13^2 = 12^2 + 4^2 + 2^2 + 2^2 + 1^2$ etc up to $13^2 = 3^2 + 2^2 + 2^2 + 1^2 + .. + 1^2$.

• do you include $0^2$ as a legal summand? – lab bhattacharjee Jul 7 '13 at 5:30
• @lab bhattacharjee: $3^2$ can not be written as the sum of two squares. Same for $7^2$ and $11^2$. – Johannes Jul 7 '13 at 5:30
• Or $17^2=16^2+4^2+4^2+1^2$. But do you get all the possible later values? – Thomas Andrews Jul 7 '13 at 6:01
• A quick thing to note is that if $n^2=a^2+b^2$ with $a,b\neq 0$, then $n$ can be written in the form $n=d(u^2+v^2)$ where $d,u,v\neq 0$. But then $$n^2=d^2\left(u^4 + 2u^2v^2 + v^2\right) = (du^2)^2 +(duv)^2 + (duv)^2 + (dv)^2$$ So $n^2$ is the sum of four non-zero squares. So you get four non-zero squares if you had two non-zero squares. – Thomas Andrews Jul 7 '13 at 6:03
• Also, if $n^2=a^2+b^2$ then (at least) one of $a,b$ is even, and then we have (say) $n^2=4a_0^2+b^2$, so we get five. Not sure how to get six. – Thomas Andrews Jul 7 '13 at 6:24

It is true.

All positive integers that are not the sum of four nonzero squares are known, see FOUR and LAGRANGE. These are the eight odd number $1,3,5,9,11,17,29,41$ and the infinite families $2 \cdot 4^t, \; \; 6 \cdot 4^t, \; \; 14 \cdot 4^t.$ Out of the list, the squares are $1,9;$ of course $1$ is not the sum of two or three nonzero squares, while $9=4+4+1$ but is not the sum of two nonzero squares. This gives a quick and dirty way to show 1,2,3 nonzero squares implies 4 nonzero squares. A more elegant method, by Thomas Andrews, is in comment above.

All numbers, square or not, larger than 33 are the sum of five nonzero squares, see FIVE. All numbers, square or not, larger than 19 are the sum of six nonzero squares, see SIX.

Now, we have some $n^2 \geq 36,$ and we are told that $n^2$ is also the sum of two nonzero squares and of three nonzero squares. ThomasAndrews shows above how $n^2$ is the sum of four nonzero squares. So far, we have 1,2,3,4,5,6 covered.

Now, take any integer $M$ with $20 \leq M < n^2.$ We are told that $M$ is the sum of six nonzero squares. Then add in $n^2 - M$ copies of 1. The result is a summation of $n^2$ as $n^2 - M + 6$ nonzero squares. Here $$6 < n^2 - M + 6 \leq n^2 - 14.$$

The numbers up to 100,000 that are nonzero squares, and are also the sum of two nonzero squares and of three nonzero squares are

         169 = 13^2
225 = 3^2 * 5^2
289 = 17^2
625 = 5^4
676 = 2^2 * 13^2
841 = 29^2
900 = 2^2 * 3^2 * 5^2
1156 = 2^2 * 17^2
1225 = 5^2 * 7^2
1369 = 37^2
1521 = 3^2 * 13^2
1681 = 41^2
2025 = 3^4 * 5^2
2500 = 2^2 * 5^4
2601 = 3^2 * 17^2
2704 = 2^4 * 13^2
2809 = 53^2
3025 = 5^2 * 11^2
3364 = 2^2 * 29^2
3600 = 2^4 * 3^2 * 5^2
3721 = 61^2
4225 = 5^2 * 13^2
4624 = 2^4 * 17^2
4900 = 2^2 * 5^2 * 7^2
5329 = 73^2
5476 = 2^2 * 37^2
5625 = 3^2 * 5^4
6084 = 2^2 * 3^2 * 13^2
6724 = 2^2 * 41^2
7225 = 5^2 * 17^2
7569 = 3^2 * 29^2
7921 = 89^2
8100 = 2^2 * 3^4 * 5^2
8281 = 7^2 * 13^2
9025 = 5^2 * 19^2
9409 = 97^2
10000 = 2^4 * 5^4
10201 = 101^2
10404 = 2^2 * 3^2 * 17^2
10816 = 2^6 * 13^2
11025 = 3^2 * 5^2 * 7^2
11236 = 2^2 * 53^2
11881 = 109^2
12100 = 2^2 * 5^2 * 11^2
12321 = 3^2 * 37^2
12769 = 113^2
13225 = 5^2 * 23^2
13456 = 2^4 * 29^2
13689 = 3^4 * 13^2
14161 = 7^2 * 17^2
14400 = 2^6 * 3^2 * 5^2
14884 = 2^2 * 61^2
15129 = 3^2 * 41^2
15625 = 5^6
16900 = 2^2 * 5^2 * 13^2
18225 = 3^6 * 5^2
18496 = 2^6 * 17^2
18769 = 137^2
19600 = 2^4 * 5^2 * 7^2
20449 = 11^2 * 13^2
21025 = 5^2 * 29^2
21316 = 2^2 * 73^2
21904 = 2^4 * 37^2
22201 = 149^2
22500 = 2^2 * 3^2 * 5^4
23409 = 3^4 * 17^2
24025 = 5^2 * 31^2
24336 = 2^4 * 3^2 * 13^2
24649 = 157^2
25281 = 3^2 * 53^2
26896 = 2^4 * 41^2
27225 = 3^2 * 5^2 * 11^2
28561 = 13^4
28900 = 2^2 * 5^2 * 17^2
29929 = 173^2
30276 = 2^2 * 3^2 * 29^2
30625 = 5^4 * 7^2
31684 = 2^2 * 89^2
32400 = 2^4 * 3^4 * 5^2
32761 = 181^2
33124 = 2^2 * 7^2 * 13^2
33489 = 3^2 * 61^2
34225 = 5^2 * 37^2
34969 = 11^2 * 17^2
36100 = 2^2 * 5^2 * 19^2
37249 = 193^2
37636 = 2^2 * 97^2
38025 = 3^2 * 5^2 * 13^2
38809 = 197^2
40000 = 2^6 * 5^4
40804 = 2^2 * 101^2
41209 = 7^2 * 29^2
41616 = 2^4 * 3^2 * 17^2
42025 = 5^2 * 41^2
43264 = 2^8 * 13^2
44100 = 2^2 * 3^2 * 5^2 * 7^2
44944 = 2^4 * 53^2
46225 = 5^2 * 43^2
47524 = 2^2 * 109^2
47961 = 3^2 * 73^2
48400 = 2^4 * 5^2 * 11^2
48841 = 13^2 * 17^2
49284 = 2^2 * 3^2 * 37^2
50625 = 3^4 * 5^4
51076 = 2^2 * 113^2
52441 = 229^2
52900 = 2^2 * 5^2 * 23^2
53824 = 2^6 * 29^2
54289 = 233^2
54756 = 2^2 * 3^4 * 13^2
55225 = 5^2 * 47^2
56644 = 2^2 * 7^2 * 17^2
57600 = 2^8 * 3^2 * 5^2
58081 = 241^2
59536 = 2^4 * 61^2
60025 = 5^2 * 7^4
60516 = 2^2 * 3^2 * 41^2
61009 = 13^2 * 19^2
62500 = 2^2 * 5^6
65025 = 3^2 * 5^2 * 17^2
66049 = 257^2
67081 = 7^2 * 37^2
67600 = 2^4 * 5^2 * 13^2
68121 = 3^4 * 29^2
70225 = 5^2 * 53^2
71289 = 3^2 * 89^2
72361 = 269^2
72900 = 2^2 * 3^6 * 5^2
73984 = 2^8 * 17^2
74529 = 3^2 * 7^2 * 13^2
75076 = 2^2 * 137^2
75625 = 5^4 * 11^2
76729 = 277^2
78400 = 2^6 * 5^2 * 7^2
78961 = 281^2
81225 = 3^2 * 5^2 * 19^2
81796 = 2^2 * 11^2 * 13^2
82369 = 7^2 * 41^2
83521 = 17^4
84100 = 2^2 * 5^2 * 29^2
84681 = 3^2 * 97^2
85264 = 2^4 * 73^2
85849 = 293^2
87025 = 5^2 * 59^2
87616 = 2^6 * 37^2
88804 = 2^2 * 149^2
89401 = 13^2 * 23^2
90000 = 2^4 * 3^2 * 5^4
91809 = 3^2 * 101^2
93025 = 5^2 * 61^2
93636 = 2^2 * 3^4 * 17^2
96100 = 2^2 * 5^2 * 31^2
97344 = 2^6 * 3^2 * 13^2
97969 = 313^2
98596 = 2^2 * 157^2
99225 = 3^4 * 5^2 * 7^2
jagy@phobeusjunior:~$ • Nice. Didn't know the five and six theorems. – Thomas Andrews Jul 7 '13 at 7:26 • @ThomasAndrews, took that out, then corrected the lower bounds for 5 and 6, which i had incorrectly switched. – Will Jagy Jul 7 '13 at 7:30 • Very nice, thank you. (You guys are quick!) – Johannes Jul 7 '13 at 8:57 • I think he is insisting on nonzero squares. – Will Jagy Jul 7 '13 at 5:29 • It is still true by induction, right? ... (Oh, but I didn't handle the upper limit...) – Dan Brumleve Jul 7 '13 at 5:30 • It's not clear how to ensure non-zero values via the induction, however. @DanBrumleve It seems like you should be able to, but there is some trickiness involved. – Thomas Andrews Jul 7 '13 at 5:36 • A summand of the form$8n+7\$ will require four non-zero squares – Mark Bennet Jul 7 '13 at 8:06