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 $m^2.$
