6
$\begingroup$

Euler famously showed that there are at least 65 idoneal (convenient) numbers. This was Euler's definition of idoneal number:

Number $n$ is idoneal if following holds: Let $m>1$ be an odd number relatively prime to n which can be written in the form $x^2+ny^2$ with $x,y$ relatively prime. If the equation $m = x^2 + ny^2$ has only one solution with $x,y\ge0$, then $m$ is a prime number.

How did Euler prove that for example $15$ or $168$, or any other, is in fact idoneal?

I am not interested in proof with Gauss's genus theory, or anything sophisticated. I am interested in techniques that were available to Euler.

$\endgroup$
2
  • $\begingroup$ Your definition is unclear. You should make it clear that $n$ is idoneal if it satisfies that property. Right now you don't use the word idoneal anywhere in the definition. $\endgroup$
    – jgon
    Dec 16 '18 at 14:51
  • $\begingroup$ Sorry. Yes, we say that $n$ is idoneal if satisfies that property. $\endgroup$ Dec 16 '18 at 15:14
2
$\begingroup$

The basic idea, as Fueter explains here (p. 19-20), is showing that for each number $m$ that is not idoneal there exists a composite number smaller than $4m$ that has a unique representation in the form $x^2 + my^2$. In modern terms, this corresponds to the result that each ideal class (or ring class when working orders) in ${\mathbb Q}(\sqrt{m})$ contains an element of norm $< 4m$. As Weil writes, Euler's presentation of this topic is highly confusing, so even he did not dare to spell out the gaps in Euler's proof.

$\endgroup$
1
  • $\begingroup$ This may indeed be true of each non-idoneal $m$, but it's true of some idoneal $m$, too. E.g. $m=2, n=6=x^2+my^2$ only if $x=2, y=1$. I take it $x\geqslant 0, y\geqslant 0$. $\endgroup$
    – Rosie F
    Feb 3 '19 at 13:57

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.