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, 2018 at 14:51
  • $\begingroup$ Sorry. Yes, we say that $n$ is idoneal if satisfies that property. $\endgroup$ Dec 16, 2018 at 15:14

1 Answer 1

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, 2019 at 13:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .