16

Happy new year! I saw a post on Facebook that surprised me:

$$2025=(1+2+3+4+5+6+7+8+9)^2=1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3+9^3$$

So, I wonder, does the equation $(a_1 + \cdots +a_9)^2=a_1^3+\cdots a_9^3$ have any other solutions over natural numbers (without zero)?

There are infinitely many integer solutions, for instance: $a_1=1, a_2=-a_9, a_3=-a_8, a_4=-a_7, a_5=-a_6$.

EDIT:

Maxime was the first to offer another solution, and there are a few more in the comments and in the other answer.

I've managed to prove that $a_1,\cdots,a_9$ must all be at most $12$. A solution with $a_1=12$ does indeed exist.

Michał Zapała
  • 1,962
  • 6
    this might not answer your question, because i've changed the number of terms in the sequence $a_1,\dots,a_9$, but for any natural number $n$ we have $(1+\cdots+n)^2 = 1^3 + \cdots + n^3$, see here for example – tkw Jan 04 '25 at 16:24
  • 3
    See also https://math.stackexchange.com/questions/216673/sum-k-1n-a-k3-left-sum-k-1n-a-k-right2 – lhf Jan 04 '25 at 17:41
  • There is nothing special about $2025=45^2=T_{9}^2$. This is just a property of triangular number $T_{n}$ as mentioned in both tkw and lhf comments.. – user25406 Jan 04 '25 at 22:33
  • @user25406, well, years that are squares are rare: 1849, 1936, 2025, 2116 are those in the centuries around the current one. – lhf Jan 09 '25 at 22:20
  • @lhf, $(1+2+3+4)^2 = (1^3 +2^3 +3^3+4^3)=10^2$. You can write a similar equation for every triangular number, in this case $Tr=10$, the fourth triangular number. – user25406 Jan 10 '25 at 13:05

4 Answers4

13

Happy new years! Another cool fact from this one you posted is $2025 = (20+25)^2$. Anyways, the equation does have another solution over natural numbers (without zero): $$(2+2+2+2+2+2+2+2 + 8)^2 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 8^3$$

Maxime Jaccon
  • 2,770
10

To summarize previous answers and comments, there are $226$ solutions (listed below) of the equation, for which $a_1 \leqslant a_2 \leqslant \cdots \leqslant a_9$.

Among them, $12$ solutions are with $2025$-property:

$\small{2025 = (1+1+2+6+6+6+7+8+8)^2 = 1^3+1^3+2^3+6^3+6^3+6^3+7^3+8^3+8^3}$ $\small{2025 = (1+1+4+5+5+5+7+8+9)^2 = 1^3+1^3+4^3+5^3+5^3+5^3+7^3+8^3+9^3}$ $\small{2025 = (1+1+5+5+5+6+6+6+10)^2 = 1^3+1^3+5^3+5^3+5^3+6^3+6^3+6^3+10^3}$ $\small{2025 = (1+2+2+5+5+7+7+7+9)^2 = 1^3+2^3+2^3+5^3+5^3+7^3+7^3+7^3+9^3}$ $\small{2025 = (1+2+3+4+5+6+7+8+9)^2 = 1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3+9^3}$ $\small{2025 = (1+2+3+5+6+6+6+6+10)^2 = 1^3+2^3+3^3+5^3+6^3+6^3+6^3+6^3+10^3}$ $\small{2025 = (1+3+3+3+4+7+8+8+8)^2 = 1^3+3^3+3^3+3^3+4^3+7^3+8^3+8^3+8^3}$ $\small{2025 = (2+2+2+3+6+7+7+7+9)^2 = 2^3+2^3+2^3+3^3+6^3+7^3+7^3+7^3+9^3}$ $\small{2025 = (2+2+4+4+4+4+8+8+9)^2 = 2^3+2^3+4^3+4^3+4^3+4^3+8^3+8^3+9^3}$ $\small{2025 = (2+3+3+3+5+5+7+7+10)^2 = 2^3+3^3+3^3+3^3+5^3+5^3+7^3+7^3+10^3}$ $\small{2025 = (3+3+3+3+3+6+6+9+9)^2 = 3^3+3^3+3^3+3^3+3^3+6^3+6^3+9^3+9^3}$ $\small{2025 = (3+3+3+3+4+5+6+8+10)^2 = 3^3+3^3+3^3+3^3+4^3+5^3+6^3+8^3+10^3}$

There is only one solution with $a_1 < a_2 < \cdots < a_9$ (well known one: $1\: 2\: 3\: 4\: 5\: 6\: 7\: 8\: 9$).

Entire list:

1) 1 1 1 1 1 2 4 4 7
2) 1 1 1 1 1 3 5 5 7
3) 1 1 1 1 2 3 3 6 7
4) 1 1 1 1 5 5 6 7 7
5) 1 1 1 2 2 4 4 5 8
6) 1 1 1 2 3 3 3 5 8
7) 1 1 1 2 3 6 6 7 7
8) 1 1 1 2 5 6 6 6 8
9) 1 1 1 4 5 5 5 5 9
10) 1 1 1 6 6 6 7 7 8
11) 1 1 2 2 2 2 4 7 7
12) 1 1 2 2 3 4 5 7 8
13) 1 1 2 2 4 6 6 7 8
14) 1 1 2 2 6 6 6 7 8
15) 1 1 2 3 4 5 5 6 9
16) 1 1 2 4 5 5 5 7 9
17) 1 1 2 5 5 6 6 7 9
18) 1 1 2 6 6 6 7 8 8
19) 1 1 3 3 3 3 5 5 9
20) 1 1 3 3 3 4 4 6 9
21) 1 1 3 3 3 4 5 8 8
22) 1 1 3 3 3 5 6 6 9
23) 1 1 3 3 5 6 7 8 8
24) 1 1 3 4 5 6 7 7 9
25) 1 1 4 5 5 5 7 8 9
26) 1 1 4 5 5 6 7 8 9
27) 1 1 5 5 5 6 6 6 10
28) 1 1 5 5 5 7 7 8 9
29) 1 1 5 5 6 6 6 6 10
30) 1 1 6 6 7 7 7 8 9
31) 1 2 2 2 2 2 2 4 8
32) 1 2 2 2 2 3 7 7 7
33) 1 2 2 2 3 3 3 7 8
34) 1 2 2 2 3 5 5 5 9
35) 1 2 2 3 3 3 7 7 8
36) 1 2 2 3 3 4 6 6 9
37) 1 2 2 3 4 4 5 7 9
38) 1 2 2 3 4 5 6 7 9
39) 1 2 2 3 4 5 7 8 8
40) 1 2 2 3 6 6 6 7 9
41) 1 2 2 5 5 5 6 8 9
42) 1 2 2 5 5 7 7 7 9
43) 1 2 2 5 6 7 7 7 9
44) 1 2 3 3 3 3 3 3 9
45) 1 2 3 3 3 3 3 6 9
46) 1 2 3 3 3 4 7 8 8
47) 1 2 3 3 5 5 8 8 8
48) 1 2 3 3 7 7 7 8 8
49) 1 2 3 4 4 4 8 8 8
50) 1 2 3 4 5 5 5 5 10
51) 1 2 3 4 5 6 7 8 9
52) 1 2 3 4 6 6 7 8 9
53) 1 2 3 5 6 6 6 6 10
54) 1 2 3 5 6 7 7 8 9
55) 1 2 3 6 6 6 6 6 10
56) 1 2 3 7 7 7 8 8 8
57) 1 2 4 4 4 4 7 8 9
58) 1 2 4 4 4 6 6 6 10
59) 1 2 4 4 6 8 8 8 8
60) 1 2 4 5 6 6 6 9 9
61) 1 2 4 6 6 7 8 8 9
62) 1 2 5 5 5 6 7 9 9
63) 1 2 5 7 7 7 8 8 9
64) 1 2 6 6 6 6 6 8 10
65) 1 3 3 3 3 5 5 8 9
66) 1 3 3 3 4 5 5 5 10
67) 1 3 3 3 4 7 8 8 8
68) 1 3 3 3 5 5 6 6 10
69) 1 3 3 3 7 7 8 8 8
70) 1 3 3 4 6 6 8 8 9
71) 1 3 3 5 5 5 7 7 10
72) 1 3 3 5 5 7 8 8 9
73) 1 3 4 4 6 7 7 7 10
74) 1 3 4 5 7 7 7 9 9
75) 1 3 4 6 7 8 8 8 9
76) 1 3 5 5 5 7 7 8 10
77) 1 3 5 6 7 7 8 9 9
78) 1 3 6 6 6 6 8 8 10
79) 1 3 6 7 8 8 8 8 9
80) 1 4 4 4 4 7 7 9 9
81) 1 4 4 7 8 8 8 8 9
82) 1 4 5 5 5 6 6 9 10
83) 1 4 5 5 7 7 8 8 10
84) 1 4 5 6 6 6 6 6 11
85) 1 4 5 6 6 6 9 9 9
86) 1 4 6 6 6 8 8 8 10
87) 1 5 5 5 5 7 7 9 10
88) 1 5 5 5 7 7 9 9 9
89) 1 5 5 6 6 6 7 7 11
90) 1 5 6 6 7 8 9 9 9
91) 2 2 2 2 2 2 2 2 8
92) 2 2 2 2 2 2 2 6 8
93) 2 2 2 2 2 4 4 4 9
94) 2 2 2 2 3 3 5 6 9
95) 2 2 2 2 3 7 7 7 8
96) 2 2 2 2 4 4 4 7 9
97) 2 2 2 2 4 6 6 7 9
98) 2 2 2 3 3 3 3 8 8
99) 2 2 2 3 5 6 6 8 9
100) 2 2 2 3 6 7 7 7 9
101) 2 2 2 4 4 4 4 8 9
102) 2 2 3 3 3 6 8 8 8
103) 2 2 3 3 4 5 5 6 10
104) 2 2 3 4 4 5 5 7 10
105) 2 2 3 5 6 6 7 9 9
106) 2 2 4 4 4 4 8 8 9
107) 2 2 4 4 4 7 8 8 9
108) 2 2 4 4 6 6 6 8 10
109) 2 2 4 4 8 8 8 8 8
110) 2 2 4 5 5 6 7 8 10
111) 2 2 5 5 5 7 8 9 9
112) 2 2 5 7 7 7 7 8 10
113) 2 2 8 8 8 8 8 8 8
114) 2 3 3 3 3 3 6 8 9
115) 2 3 3 3 3 4 4 5 10
116) 2 3 3 3 3 4 5 6 10
117) 2 3 3 3 3 4 7 8 9
118) 2 3 3 3 3 6 6 6 10
119) 2 3 3 3 5 5 7 7 10
120) 2 3 3 3 5 6 7 7 10
121) 2 3 3 3 6 7 8 8 9
122) 2 3 3 4 4 5 7 9 9
123) 2 3 3 5 6 7 7 8 10
124) 2 3 4 4 5 7 8 9 9
125) 2 3 4 4 7 7 7 8 10
126) 2 3 4 5 6 6 6 9 10
127) 2 3 4 6 7 7 8 8 10
128) 2 3 5 5 5 5 5 5 11
129) 2 3 5 5 6 7 7 9 10
130) 2 3 5 6 7 7 9 9 9
131) 2 3 6 6 6 6 7 7 11
132) 2 3 7 7 7 8 8 8 10
133) 2 4 4 4 5 5 5 9 10
134) 2 4 4 4 5 6 6 6 11
135) 2 4 4 4 6 6 7 9 10
136) 2 4 4 4 7 8 8 9 9
137) 2 4 4 5 5 5 6 7 11
138) 2 4 4 5 5 7 9 9 9
139) 2 4 4 6 6 6 8 9 10
140) 2 4 5 5 7 7 7 7 11
141) 2 4 5 5 7 7 8 9 10
142) 2 4 5 5 7 8 9 9 9
143) 2 4 6 6 6 7 7 8 11
144) 2 4 6 6 7 8 8 9 10
145) 2 4 7 7 8 8 9 9 9
146) 2 5 5 5 5 5 5 8 11
147) 2 5 5 6 6 6 7 10 10
148) 2 5 5 6 6 6 9 9 10
149) 2 5 6 7 7 7 7 10 10
150) 2 6 6 6 6 6 6 9 11
151) 2 6 6 6 6 6 8 10 10
152) 2 6 6 6 7 8 9 9 10
153) 2 6 6 7 7 8 9 9 10
154) 3 3 3 3 3 6 6 9 9
155) 3 3 3 3 4 5 6 8 10
156) 3 3 3 3 4 6 6 8 10
157) 3 3 3 3 5 6 7 8 10
158) 3 3 3 4 4 4 4 8 10
159) 3 3 3 4 4 5 8 9 9
160) 3 3 3 4 6 6 8 8 10
161) 3 3 3 5 5 5 6 9 10
162) 3 3 3 5 6 6 6 6 11
163) 3 3 3 5 7 7 8 8 10
164) 3 3 3 6 6 6 9 9 9
165) 3 3 3 7 7 8 8 9 9
166) 3 3 4 5 5 6 7 7 11
167) 3 3 4 5 6 6 8 9 10
168) 3 3 4 6 6 8 9 9 9
169) 3 3 5 5 5 6 6 8 11
170) 3 3 5 6 6 8 8 9 10
171) 3 4 4 4 5 7 7 7 11
172) 3 4 4 5 6 7 7 8 11
173) 3 4 4 5 6 8 8 9 10
174) 3 4 5 5 5 6 7 10 10
175) 3 4 5 5 7 9 9 9 9
176) 3 4 5 7 7 7 8 8 11
177) 3 4 6 6 7 7 8 10 10
178) 3 5 5 5 6 7 7 9 11
179) 3 5 5 5 7 7 8 10 10
180) 3 5 6 7 7 7 8 9 11
181) 3 5 6 7 9 9 9 9 9
182) 3 5 7 7 7 7 8 9 11
183) 3 5 7 8 8 8 9 9 10
184) 3 6 7 7 7 8 8 9 11
185) 3 7 7 8 8 8 8 10 10
186) 4 4 4 4 4 4 7 7 11
187) 4 4 4 5 5 5 9 9 10
188) 4 4 4 5 6 6 6 9 11
189) 4 4 4 6 7 8 9 9 10
190) 4 4 4 8 8 8 8 9 10
191) 4 4 5 5 5 7 7 9 11
192) 4 4 5 6 6 7 8 9 11
193) 4 4 6 6 6 6 9 10 10
194) 4 4 6 6 8 8 8 10 10
195) 4 5 5 5 6 6 9 10 10
196) 4 5 5 5 7 9 9 9 10
197) 4 5 5 6 6 6 6 7 12
198) 4 5 5 7 8 9 9 9 10
199) 4 5 6 6 6 6 6 10 11
200) 4 5 6 6 6 7 9 9 11
201) 4 5 6 6 7 7 7 7 12
202) 4 5 6 6 7 7 9 9 11
203) 4 6 6 6 6 6 6 8 12
204) 4 6 6 6 7 8 9 9 11
205) 4 6 6 8 8 8 9 10 10
206) 4 6 8 8 8 8 8 9 11
207) 4 7 8 8 8 8 9 10 10
208) 5 5 5 5 6 6 9 9 11
209) 5 5 5 5 7 7 7 7 12
210) 5 5 5 5 8 8 8 9 11
211) 5 5 6 6 6 9 9 10 10
212) 5 6 6 7 7 9 9 9 11
213) 5 6 7 9 9 9 9 9 10
214) 5 7 7 7 7 7 9 10 11
215) 5 7 7 7 7 8 8 8 12
216) 5 7 7 7 8 9 9 9 11
217) 5 7 8 9 9 9 9 9 10
218) 6 6 6 6 6 6 6 9 12
219) 6 6 6 6 6 6 7 9 12
220) 6 6 6 6 6 6 9 10 11
221) 6 6 6 8 8 8 10 10 10
222) 6 7 7 7 7 7 8 9 12
223) 6 7 7 7 8 8 9 10 11
224) 6 8 8 8 8 8 10 10 10
225) 7 7 7 8 8 8 9 10 11
226) 9 9 9 9 9 9 9 9 9
Oleg567
  • 17,740
  • Out of curiosity, do you have any idea about the sharp bound of $M=\max(a_i)$. As numerical evidence shows $M\leq 12$ but how do we find that $12$. – Quý Nhân Jan 05 '25 at 11:02
  • +1 Did you find these solutions by a brute force computer search, or by some other method? – Adam Bailey Jan 05 '25 at 11:05
  • Quy Nhan, I suspect $12$ comes from $(25-1)/2=12$. – user25406 Jan 05 '25 at 11:11
  • @Quý Nhân : If $a_1 \leqslant a_2 \leqslant a_9$, then denote $x_j = a_j/a_9$, then $$ a_9^2(x_1+\ldots+x_8+1)^2 = a_9^3(x_1^3+\ldots+x_8^3+1) $$ $$ a_9^2(s_1+1)^2 = a_9^3(s_3+1), $$ where $s_1 = x_1+\ldots+x_8$, $s_3 = x_1^3+\ldots+x_8^3$. $$ (s_1+1)^2 = a_9(s_3+1). $$ $$ s_3+1 = (s_1+1)^2/a_9.;;(*) $$ ... – Oleg567 Jan 05 '25 at 12:56
  • 2
    ...

    By https://en.wikipedia.org/wiki/Generalized_mean#Generalized_mean_inequality, $$ \frac{s_1}{8} \leqslant \frac{s_3^{1/3}}{8^{1/3}} $$ $$ \left(\frac{s_1}{8}\right)^3 \leqslant \frac{s_3}{8} $$

    $$ \frac{s_1^3}{64} \leqslant s_3 $$ from (*) $$ \frac{s_1^3}{64} \leqslant (s_1+1)^2/a_9 - 1 $$ $$ 1+\frac{s_1^3}{64} \leqslant (s_1+1)^2/a_9 $$

    $$ a_9 \leqslant \frac{(s_1+1)^2}{1+\frac{s_1^3}{64}}, $$ and max value of RHS for $s_1 \in (0, 8]$ is $\approx 12.53$.

     – Oleg567 Jan 05 '25 at 12:56
  • @Adam Bailey: yes, I used simple brute force computer search. – Oleg567 Jan 05 '25 at 12:57
  • @Oleg567 : by the same argument, it can be showed that a generalized version with $n$ variables is bounded at roughly $0.53 \times n^\frac43$ – Michał Zapała Jan 05 '25 at 14:40
  • @Oleg567 Welcome back. I think I sent you a msg about a post last year. Can you check your inbox? – Tito Piezas III Jan 05 '25 at 16:53
9

Another solution, obtained by trial and error:

$(1+1+1+2+2+4+4+5+8)^2=1^3+1^3+1^3+2^3+2^3+4^3+4^3+5^3+8^3$

Adam Bailey
  • 4,735
3

A bit off from OP's question, consider Diophantine equation: $$\sum_{k=1}^{n}a_{k}^{3}=\left(\sum_{k=1}^{n}a_{k}\right)^2,\ a_k\in\mathbb{Z}_{+}$$ I will generalize Oleg567's comment. And from now on, I will work with $n\geq 3$.
Synopsis: using inequality $\textrm{average}(x_{j})\leq \textrm{average}(x_{j}^{3})^{1/3}$, we can prove $\max(a_{k})\leq\frac{(s_1+1)^2}{1+\frac{s_1^3}{(n-1)^2}},s_1\in(0,n-1]$ where $s_1=-1+\frac{1}{\max(a_k)}\sum_{j=1}^n a_{j}$.

From there, with basic calculus and CAS, it can be shown that: $$\max(a_k)\leq M(n)=\left\lfloor \frac{(\sqrt[3]{b^2}+\sqrt[3]{2})(2(n-1)\sqrt[3]{b}+\sqrt[3]{2b^2}+\sqrt[3]{4})}{3\sqrt[3]{4b^2}}\right\rfloor $$ $$ b=n-1+\sqrt{n^2-2n-1}$$ Consequently: $M(n)\sim \frac{\sqrt[3]{4}}{3}n^{4/3}$

Quý Nhân
  • 2,706
  • Another interesting question is how small $\sigma = \sum_{i=1}^n a_i$ is allowed to be. A lower bound is provided by the condition $(\sigma-n+1)^3+n-1\ge \sigma^2$. This is approximately $n+O(n^\frac23)$, but maybe one can do better. An upper bound is obviously $n^2$, and it's always possible to make it with $a_1 = \cdots = a_n = n$. – Michał Zapała Jan 05 '25 at 15:57
  • @MichałZapała I don't have a proof but it happens when $n-1$ entries are $1$, Mathematica gives $$\min\sigma \sim n+n^{2/3}+\frac{2n^{1/3}}{3}-\frac{2}{3}-\frac{71}{81 n^{1/3}}+\frac{7}{243 n^{2/3}}-\frac{2047}{6561 n^{4/3}}+\frac{4591}{19683 n^{5/3}}$$ – Quý Nhân Jan 13 '25 at 13:53