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.

How can I show that the equation $$3x^3 + 4y^3 + 5z^3 = 0$$ has nonzero solutions modulo every integer but not in $\mathbb{Z}$?

share|improve this question

1 Answer 1

up vote 13 down vote accepted

This is, of course, Selmer's famous example that the Hasse principle breaks down for cubic (or higher) forms.

It is easy to come up with solutions modulo any power of $2$, any power of $3$, and any power of $5$. For larger primes, if $p\equiv 2\pmod{3}$ then every nonzero element modulo $p$ is a cube, so picking your favorite values for $x$ and $y$ yields a solution $z$, and then you can use Hensel's Lemma to lift that solution (as a cubic polynomial in $z$) modulo $p^n$ for any $n$. If $p\equiv 1\pmod{3}$, then you just need to find a pair $x$ and $y$, not both zero, such that $3x^3+4y^3$ and $-5$ have the same cubic character modulo $p$, and that gives a solution for $z$; Hensel's Lemma again lets you lift it modulo $p^n$ for all $n$.

From knowing that it has nonzero solutions modulo every prime power you conclude using the Chinese Remainder Theorem that it has solutions modulo every integer.

Showing that it has no integer solutions is, as far as I know, a bit harder. One method I seem to recall relies on multiplying through by $2$, and factoring $6X^3 + Y^3 = 10Z^3$ over $\mathbb{Q}(\sqrt[3]{6})$ to show there are no solutions in $\mathbb{Z}$.

Added. A web search reveals a handout by Keith Conrad that contains the proof in detail, using $p$-adic numbers and arithmetic in $\mathbb{Q}(\sqrt[3]{6})$.

share|improve this answer
    
Shouldn't it be $-25(3x^3+4y^3)$? And why is it obvious that we can find such $x$ and $y$? That's what seemed like the difficult part to me. –  joriki Aug 2 '11 at 12:59
    
In the handout, Keith Conrad uses an inequality derived using character sums over finite fields to show there's a solution for $p\equiv 1\pmod{3}$, so it seems it can't be entirely trivial? :-) –  joriki Aug 2 '11 at 13:21
    
@joriki: I had that wrong; you want $(3x^3+4y^3)$ to have the same cubic character as $-5$; then they differ by a cube, so $3x^3+4y^3\equiv -5z^3\pmod{p}$ will have a nonzero solution $z$. –  Arturo Magidin Aug 2 '11 at 14:35
    
@joriki: Well, I swept the difficulty under the rug by saying you "just" need to find a pair $x$ and $y$, both nonzero, such that $3x^3+4y^3$ has the same cubic character as $-5$; how can you guarantee such a pair exists? –  Arturo Magidin Aug 2 '11 at 14:36

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.