# 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

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Let $1^2+2^2+\cdots+k^2=m$ and multiply through by $m^2.$
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 at 4:54