<s>[INST] How prove this diophantine equation $x^2+y^2+z^3=n$ always have integer solution 
show that: For any postive ineteger $n$,then the equation 
  $$n=x^2+y^2+z^3$$
  always have integer solution

My idea: such as $n=1$,then we have
$$1=0^2+0^2+1^3$$
$$2=0^2+1^2+1^3$$
$$3=1^2+1^2+1^3$$
$$4=2^2+0^2+0^3$$
$$5=1^2+2^2+0^3$$
$$6=1^2+2^2+1^3$$
$$7=2^2+2^2+(-1)^3$$
$$8=0^2+0^2+2^3$$
$$9=1^2+0^2+2^3$$
$$10=1^2+1^2+2^3$$
$\cdots\cdots\cdots$
But for  general $n$, How prove it?
 [/INST] My friend put this as an MAA Monthly problem, years ago. The comparison is that there are infinitely many numbers that have no expression as $x^2 + y^2 + z^9.$ This simple result defeated an existing conjecture; we sent it early to Robert C. Vaughan, so it made it into the second edition of his book The Hardy-Littlewood Method. It is likely that every number can be written as $x^2 + y^2 + z^5,$ but not certain. 
See http://zakuski.utsa.edu/~jagy/Elkies_Kap.pdf
and related items at http://zakuski.utsa.edu/~jagy/inhom.html
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