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Find a nontrivial answer for the Diophantine system below:

$$a_1+a_2+...+a_m=b_1+b_2+...+b_n$$ $$a_1^2+a_2^2+...+a_m^2=b_1^2+b_2^2+...+b_n^2$$ $$a_1^3+a_2^3+...+a_m^3=b_1^3+b_2^3+...+b_n^3$$

where $a_1,...a_m$ and $b_1,...,b_n$ are all natural numbers.

my attempt:

I tried defactorizing all variables to primes for at least a special case but it led me to no way...I think letting variables to be negative may simplify the problem but the case in which all variables are natural seems more tricky...

Currently I have no idea where to start over and I appreciate any set of answers....

Thanks in advance!

Mostafa Ayaz
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  • $1+4=2+3$, $1^2+4^2+5^2+8^2 =2^2+3^2+6^2+7^2$, you can do similar things to arbitrarily large powers. In particular for your cubic case there is a solution with 8 numbers on each side. – Hw Chu Jan 05 '18 at 21:23
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    Please read Prouhet-Tarry-Escott problem Wikipedia article. – Somos Jan 05 '18 at 22:34
  • The problem should be formulated differently. If you want to obtain a formula for some kind of system. Need the system itself to write. Actually the formula is similar, but for each system with its number of summands has its own formula. – individ Jan 06 '18 at 04:41
  • Dear Hw Chu!....Unfortunately $\lbrace{1,4}\rbrace\ne\lbrace{1,4,5,8}\rbrace$ and $\lbrace{2,3}\rbrace\ne\lbrace{2,3,6,7}\rbrace$ – Mostafa Ayaz Jan 06 '18 at 08:59

1 Answers1

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$$\left\{\begin{aligned}&a+b+c+d=x+y+z+v\\&a^2+b^2+c^2+d^2=x^2+y^2+z^2+v^2\\&a^3+b^3+c^3+d^3=x^3+y^3+z^3+v^3\end{aligned}\right.$$

The solution of the system of equations can be written in this form.

$$a=(t^2+k^2)p-(t^2-k^2)s+(t-k)kr+Q$$

$$b=(t^2+k^2)p-(t^2-k^2)s+(t+k)kr+Q$$

$$c=2sk^2+Q$$

$$d=2((t^2+k^2)p-st^2+tkr)+Q$$

$$x=(t^2+2tk+k^2)p-(t^2+2tk-k^2)s+(t+k)kr+Q$$

$$y=(t^2-2tk+k^2)p-(t^2-2tk-k^2)s+(t-k)kr+Q$$

$$z=2pk^2+Q$$

$$v=2(pt^2-(t^2-k^2)s+tkr)+Q$$

$t,k,p,s,r,Q - $ are any integer.

individ
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