3

The Jacobi-Madden equation $ a^4+b^4+c^4+d^4 = (a+b+c+d)^4 $ has an infinitude of integer solutions with all variables non-zero.

Is it known if the same is true for the general equation $ \sum\limits_{i=1}^n a_i^n = (\sum\limits_{i=1}^n a_i)^n$ ?

I tried to look for some references, but found nothing.

user967210
  • 711

2 Answers2

4

There is a similar question I answered back in 2014. See the MSE post "Integer Solutions of $x^3+y^3+z^3 = (x+y+z)^3.\,$ In summary,

I. 4th powers

There are infinitely many solutions to,

$$a^4+b^4+c^4+d^4=(a+b+c+d)^4$$

given by the Jacobi-Madden equation (2008).

II. 5th powers

There are also infinitely many solutions to,

$$a^5+b^5+c^5+d^5+e^5=(a+b+c+d+e)^5$$

given by Lander (1968). See "Geometric Aspects of Diophantine equations involving Equal Sums of Like Powers".

Note: Choudhry, Wroblewski, and others would also give more examples.

III. 6th powers

In Meyrignac's database http://euler.free.fr/database.txt, there are exactly 177+1 = 178 solutions to,

$$x_1^6+x_2^6+x_3^6+x_4^6+x_5^6+x_6^6+x_7^6 = z^6$$

mostly found between 2009 to 2012. Peter Ansell and others found many of these (6,1,7) and the aim was to find a (6,1,6). Ansell discovered four with the additional property,

$$x_1^6+x_2^6+x_3^6+\dots+x_7^6 = (x_1+x_2+x_3+\dots+x_7)^6$$

But he may have tested only his solutions, so there could be more.

IV. 7th powers

Choudhry and Wroblewski gave more than 20 solutions to the (7,1,7) that is also,

$$x_1^7+x_2^7+x_3^7+\dots+x_7^7 = (x_1+x_2+x_3+\dots+x_7)^7$$

See this MO link.

V. 8th powers

There is a (8,1,8) and a (8,1,9), but only the (8,1,10) satisfies,

$$x_1^8+x_2^8+x_3^8+\dots+x_{10}^8\, = \,(x_1+x_2+x_3+\dots+x_{10})^8$$

with $x_i$ as $(6, 16, 34, 58, 66, 142, 152, 171, -184, -226)$.

VI. 9th powers

There are many solutions to (9,1,9) but Wroblewski found the only one that is also,

$$x_1^9+x_2^9+x_3^9+\dots+x_9^9 = (x_1+x_2+x_3+\dots+x_9)^9$$

with $x_i$ as $(-100, -187, -429, -603, -621, 51, 253, 412, 600)$.

Note: Bremner and Delorme ("On Equal Sums of Ninth Powers") found infinitely many to,

$$x_1^9+x_2^9+x_3^9+\dots+x_{11}^9 = (x_1+x_2+x_3+\dots+x_{11})^9$$

VII. 10th powers

  1. A similar result uses a (10,1,12) by Wroblewski,

$$x_1^{10}+x_2^{10}+x_3^{10}+\dots+x_{12}^{10} = 2^{10}(x_1+x_2+x_3+\dots+x_{12})^{10}$$

The $x_i$ are $(-62, -115, 172, 245, -295, -533, 689, 927, 1011, -1234, -1603, 1684).$

  1. Another uses a (10,1,13) by S. Chase,

$$x_1^{10}+x_2^{10}+x_3^{10}+\dots+x_{13}^{10} = (x_1+x_2+x_3+\dots+x_{13})^{10}$$

The $x_i$ are $(-187, -179, -128, -59, -13, -6, 49, 57, 73, 85, 122, 204, 210).$

Tito Piezas III
  • 54,565
1

In Peter Ansell's old website, he has given three four numerical solutions to the equation:

$$a^6+b^6+c^6+d^6+e^6+f^6+g^6=(a+b+c+d+e+f+g)^6=z^6$$

where $(a,b,c,d,e,f,g;z)$ are,

$$(3936,4230,3300,1854,-888,-3187,-4170;\, 5075)\\(3912,3858,2654,3300,-495,-3018,-4884;\, 5327)\\(3546,7209,9194,2046,108,-2712,-8982;\, 10409)\\ (20, 4260, 5778, 11157, 12414, - 7278, - 11994;\, 14357)$$

Tito Piezas III
  • 54,565
David
  • 71
  • Thank you! Could you give me a reference? – user967210 Aug 08 '23 at 03:31
  • Arriving at a formula & finding new numerical solutions from the above answer for degree six could be difficult. It was possible for degree four. For degree three there is a solution given by Ramanujan for (2a,2b,2c)^3=(a+b+c)^3 where [2a=6x^2+10xy-10y^2),2b=(8x^2-8xy+12y^2),2c=(10x^2-10xy-6y^2)]. For y=0 we get (6,8,10)^3=(3+4+5)^3=(12)^3 – David Aug 09 '23 at 06:11
  • Cont'd: I noticed that there is parametrization for degree two. (a^2+b^2+c^2)=(a-b+c)^2 where (a,b,c)=(3p,3q,pq) & p=(3k-2) & q=(3k-5) & for k=2, we get (a,b,c)=(12,4,-3). Also, for degree five there is an interesting numerical solution: (a,b,c,d,e)^5=(4(a-b)+d+e+f)^5=(w)^5 where (a,b,c,d,e,w)=(100,80,57,43,7,107) – David Aug 10 '23 at 14:38
  • @user967210 See http://euler.free.fr/database.txt and search for the term "Ansell". – Tito Piezas III Aug 13 '23 at 05:36