Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It's easy$^*$ to prove that if $n=3^{6m}(3k \pm 1)$ where $(m,k) \in \mathbb{N} \times \mathbb{Z}$, then $n=a^2+b^2+c^3+d^3$ with $(a,b,c,d) \in \mathbb{Z}^4$. But how to prove that this is true if $n=3k$?

Thanks,

W

$^*$ Because $3k+1=0^2+(3k+8)^2+(k+1)^3+(-k-4)^3$, $3k+2=1^2+(3k+8)^2+(k+1)^3+(-k-4)^3$ and $3^6(a^2+b^2+c^3+d^3)=(27a)^2+(27b)^2+(9c)^3+(9d)^3$.

share|improve this question
    
What is $\mathbb Z^4$? –  Jorge Fernández Jul 23 '12 at 13:56
4  
@ChuckFernández I presume the set of all 4-tuples comprised of integers? –  Shaktal Jul 23 '12 at 14:06
    
does it mean the same as $a,b,c,d\in \mathbb Z$? –  Jorge Fernández Jul 23 '12 at 15:00
    
@ChuckFernández Yes. –  Ragib Zaman Jul 23 '12 at 15:19

1 Answer 1

You do not need the final $d^3.$ Every integer is the sum of two squares and a cube, as long as we do not restrict the $\pm$ sign on the cube. TYPESET FOR LEGIBILITY:

Solution by Andrew Adler: $$ 2x+1 = (x^3 - 3 x^2 + x)^2 +(x^2 - x - 1)^2 -(x^2 - 2x)^3 $$ $$ 4x+2 = (2x^3 - 2 x^2 - x)^2 +(2x^3 -4x^2 - x + 1)^2 -(2x^2 - 2x-1)^3 $$ $$ 8x+4 = (x^3 + x +2 )^2 +(x^2 - 2x - 1)^2 -(x^2 + 1)^3 $$ $$ 16x+8 = (2x^3 - 8 x^2 +4 x +2)^2 +(2x^3 -4x^2 - 2 )^2 -(2x^2 - 4x)^3 $$ $$ 16x = (x^3 +7 x - 2)^2 +(x^2 +2 x + 11)^2 -(x^2 +5)^3 $$

You can check these with your own computer algebra system. Please let me know if I mistyped anything.

Alright, our conjecture (Kaplansky and I) is that, for any odd prime $q,$ $x^2 + y^2 + z^q$ is universal. However, this is false as soon as the exponent on $z$ is odd but composite. The example we put in the article is $$ x^2 + y^2 + z^9 \neq 216 p^3, $$ where $p \equiv 1 \pmod 4$ is a (positive) prime. This defeated a well-known conjecture of Vaughan. We told him about it in time for him to include it in the second edition of his HARDY-LITTLEWOOD BOOK, where it is now mentioned on pages 127 ("There are some exceptions to this,") and exercise 5 on page 146.

enter image description here

share|improve this answer
    
Looks like one could crop out 3/4th of the picture. (Also, $\div$!?) –  anon Jul 23 '12 at 19:35
    
@anon , I have no idea how to do such things whatsoever. If anyone knowledgeable wants to fiddle with this, make sure we can see the page numbers and issue number and years. It is the American Mathematical Monthly published by the M.A.A. I was going to typeset the five equations in the answer by Adler, as they seem close to illegible on my screen. –  Will Jagy Jul 23 '12 at 19:41
    
@Will Jagy : Thank you very much ! –  Watsoon Jul 23 '12 at 20:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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