# How prove for any $k$ then have $a^2_{1}+a^2_{2}+a^2_{3}+\cdots+a^2_{k}=m^3$

for any positive integer $k$,there exsit $m\in N$ and $a_{i}\in N,i=1,2,\cdots,k$,such

(1): $a_{i}\neq a_{j},i\forall i\neq j$,

(2): $$a^2_{1}+a^2_{2}+a^2_{3}+\cdots+a^2_{k}=m^3$$

My idea: if $k=1$,then we let $a_{1}=1$,then $$a^2_{1}=1^3=m^3$$ if $k=2$, then we let $a_{1}=5,a_{2}=10$,then we have $$a^2_{1}+a^2_{2}=5^2+10^2=125=5^3=m^3$$ if $k=3$, note $$3^2+4^2+10^2=125=5^3$$ then let $a_{1}=3,a_{2}=4,a_{3}=10,m=5$

But follow I can't find it and How prove this problem ,Thank you

-

Let $1^2+2^2+\cdots+k^2=m$ and multiply through by $m^2.$

-
oh!,It's nice !!+1 – china math Mar 1 '14 at 4:32
One can use any set of $k$ distinct squares on the left side, then define their sum as $m$, and "multiply through by $m^2$" as above to get more examples. Question left: what are all the solutions? – coffeemath Mar 1 '14 at 4:54
@coffeemath: There is also an identity such that the $a_i$ need not have a common factor, and I have given it below. – Tito Piezas III Dec 17 '14 at 4:39
@TitoPiezasIII I looked at that identity, and it seems interesting. (Gave the answer an upvote) – coffeemath Dec 17 '14 at 4:45

(Edit: Gave a more general answer.)

Actually, it's known that you can non-trivially solve the equation,

$$a_1^2+a_2^2+a_3^2+\dots+a_n^2 = (b_1^2+b_2^2+b_3^2+\dots+b_n^2)^k\tag1$$

for any positive integer $k$, but the $a_i$ (except $a_1$) will have a common factor. If you want that the $a_i$ do not generally have a common factor, then you can do so for $k=4m+3$. For the simplest case $k=3$, we have,

$$x_1^2+x_2^2+x_3^2+\dots+x_n^2 = (y_1^2+y_2^2+y_3^2+\dots+y_n^2)^3\tag2$$

where the $x_i$ are,

$$x_1 = p y_1-q y_2,\quad x_2 = p y_2-q y_3,\quad x_3 = p y_3-q y_4,\quad\dots\quad x_n = p y_n-q y_1$$

and,

$$p = y_1^2+y_2^2+y_3^2+\dots+y_n^2$$

$$q = 2(y_1y_2+y_2y_3+y_3y_4+\dots+y_ny_1)$$

for $n$ free variables $y_i$. I trust the pattern of the $x_i$ is clear?

(There is an iterative process that generates the other $k=4m+3$, but is too tedious to explain here, so we will just give a numerical example.)

For $n=4$, let the $y_i$ be the four primes $y_i = 2,11,13,17$. So we get,

\begin{aligned} &583 = 2^2+11^2+13^2+17^2\\ &583^3 = 4507^2+8074^2+6701^2+8231^2\\ &583^7 = 2063328403^2+187823746^2+2402985509^2+3581180761^2\\ &583^{11} = 28216171649227^2+1234758982541722^2+745071137544115^2+751004146893527^2 \end{aligned}

If we swap the $y_i$, we get different addends $x_i$, but the same sum,

\begin{aligned} &583 = 2^2+11^2+17^2+13^2\\ &583^3 = 8866^2+9091^2+1945^2+5755^2\\ &583^7 = 951019982^2+798185251^2+3073066009^2+3450583061^2\\ &583^{11} = 434781267760906^2+434087631430181^2+1271549279992223^2+806386837663771^2 \end{aligned}

and so on. You can check that the addends do not have a common factor. Nice, eh?

-