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Is there a way to show that every integer $n$ can be represented by
$(a^3+b^2+c^2)-(d^3+e^2+f^2)$
Where $a,b,c,d,e,f$ are all integers?

Shuaib Lateef
  • 439

3 Answers3

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Actually, every integer can be expressed as $$ b^2 + c^2 - d^3 $$

This was a problem Kap and Noam Elkies gave to the MAA Monthly, Problems and Solutions column. It appeared in January 1995, Volume 102, number 1, page 70. The answers appeared in June-July 1997, Volume 104, number 6, page 574.

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However, not every number can be expressed as $ b^2 + c^2 - d^9 $

Much later, we found that not every integer can be expressed as $$ 4 x^2 + 2 xy + 7 y^2 - z^3 $$

I sent that one into the Monthly, the question appeared in December 2010. The single answer, by Robin Chapman, appeared in December 2012, volume 119, number 10, pages 884-885

For the curious, see the more difficult https://mathoverflow.net/questions/12486/integers-not-represented-by-2-x2-x-y-3-y2-z3-z

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Will Jagy
  • 139,541
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Yes, every integer $k$ is of this form. This is trivial, because we may take $a=d$ and some of the $b,c,e,f$ equal to zero. Every $k$ is either of the form $2n+1$ or $2n$. We have \begin{align*} 2n+1 & = (n+1)^2-n^2+0^2 \\ 2n & =(n+1)^2-n^2-1^2 \end{align*}

Dietrich Burde
  • 130,978
1

I have found that every integer $n$ can be represented by
$(a^2+b^2)-(c^2+d^2)$
Because $$n=((\frac{(n+1)(n+2)}{2})^2 + (\frac{n(n+1)(n+2)}{2})^2)-\left((\frac{n(n+1)(n+2)}{2}+1)^2 + (\frac{n(n+1)}{2})^2\right)$$
So, I am just wondering if there is at least an expression which suits the question i asked.

Dietrich Burde
  • 130,978
Shuaib Lateef
  • 439
  • Please, have a look at the order of the brackets again and you will see the equation is actually correct. I think you mistakenly thought I meant $n=a^2+b^2-c^2+d^2$ which is not but $n=a^2+b^2-c^2-d^2$. – Shuaib Lateef Apr 06 '21 at 13:40
  • I think the first term is equal to 1, don't you agree? – Shuaib Lateef Apr 06 '21 at 14:17
  • Yes, now I have seen it - I am sorry. But you can use a simpler formula, namely just $2n=(n+1)^2-n^2-1$ etc. – Dietrich Burde Apr 06 '21 at 14:19
  • Thanks for the suggestion but I am curious to know if there are expressions which satisfy $a,b,c,d,e,f$ in my original question. – Shuaib Lateef Apr 06 '21 at 14:47
  • Yes, of course there are. You have just given such an expression with $a=d=0$. My answer is even easier with, say, $a=b=d=0$. If you wish, your answer gives such an expression also for nonzero $a,d$ with $a^3=d^3$. So you don't need $a$ and $d$, right? – Dietrich Burde Apr 06 '21 at 14:49
  • Oh! I see. Thanks a lot. – Shuaib Lateef Apr 06 '21 at 14:54