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?

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    $\begingroup$ Too Broad is a particular wrong reason to close this. Is somebody really being a bad actor? $\endgroup$ – Thomas Andrews Nov 26 '14 at 3:40

Here’s a proof using a solution I found here.

Write $n$ in the form $8^m\cdot s$, for an integer $m\ge0$ and with $s$ not divisible by $8$. This can always be done. Note that an integer $s$ that is not a multiple of $8$ can be written in one of the following three forms: $2k+1$ (if $s$ is odd), $4k+2$ (if $s$ is even, but not a multiple of $4$), or $8k+4$ (if $s$ is even and a multiple of $4$).

First, find an expression for $s$ in the form $a^2+b^2+c^3$ as follows. (I haven’t checked these details.)

If $s=2k+1$, let $a=k^2-k-1$, $b=k^3-3k^2+k$, and $c=-k^2+2k$.

If $s=4k+2$, let $a=2k^3-2k^2-k$, $b=2k^3-4k^2-k+1$, and $c=-2k^2+2k+1$

If $s=8k+4$, let $a=k^2-2k-1$, $b=k^3+k+2$, and $c=-k^2-1$

Now observe that if $s$ is the sum of two squares and a cube, so is $8s$, because $8(a^2+b^2+c^3)=(2a+2b)^2+(2a-2b)^2+(2c)^3$. Inductively, if $s$ is the sum of two squares and a cube, then so is $8^ms$ for any nonnegative integer $m$.

We have already shown that $s$ is the sum of two squares and a cube, so $n=8^ms$ is as well, completing the proof.

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  • $\begingroup$ The case $s = 8k+4$ is not correct. The expression $a^2 + b^2 + c^3$ comes out to $4-4k^4$. $\endgroup$ – RghtHndSd Nov 26 '14 at 3:50
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    $\begingroup$ $b = k^3+k+2$. I've taken the liberty to edit your answer. $\endgroup$ – RghtHndSd Nov 26 '14 at 3:56
  • $\begingroup$ @RghtHndSd: Thank you! $\endgroup$ – Steve Kass Nov 26 '14 at 4:07

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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  • $\begingroup$ Nice indentity,thank you! $\endgroup$ – math110 Nov 26 '14 at 2:59

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