13
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ $b = k^3+k+2$. 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
5
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.