8
$\begingroup$

Peter Barlow was a well-known British mathematician during the 19th century and he produced the very first English number-theoretic text called "An Elementary Investigation of the Theory of Numbers" in 1811. This text contained some notable errors (e.g: an incorrect proof of Fermat's Last Theorem and several false theorems on perfect numbers) but it was still influential and was used as a reference guide.

In this text, I just discovered that he claimed to have solved the negative pell's equation and gave a proof for it. Ofcourse, he was most likely wrong since we still today don't know when it is solvable. His argument seems somewhat intricate and it would be extremely interesting to know where he went wrong.

He states the theorem as follows;

If $a$ be a prime number of the form $4n+1$, the equation $x^2-ay^2=-1$ is always resolvible in integers.

and then goes on to prove it in the following manner:

(Barlow uses the symbol "ⵐ" to mean "is of the form".)

Let $p$ and $q$ be least values (except 1 and 0) that satisfy the equation

$p^2-aq^2=1$, or $p^2-(4n+1)q^2=1$;

then it is obvious that q must be even, for if it was odd, $q^2$ would be of the form 8n+1, and

$aq^2+1$ ⵐ $(4n+1)*(8n'+1)+1$ ⵐ $4n''+2$,

which cannot be a square: since, then, q must be even, let us make $q=2mn$, $m$ and $n$ being integers prime to each other; then we shall have $p^2-1=4m^2n^2$; but $p$ being odd, and consequently, $p^2=1$ ⵐ $8n'$, it follows, that either $m$ or $n$ is even, and the other odd, for otherwise we should not have $p^2-1$ ⵐ $8n'$, and they cannot be both even, because they are prime to eachother.

Let us therefore suppose $n$ to be odd, then the equation

$(p+1)(p-1)=4am^2n^2$,

in which the factors $p+1$, and $p-1$, can have only the common measure $2$ (and this must necessarily have place, because $p$ is odd), will be resolvible into the four following forms:

  1. $p+1=2am^2$ & $p-1=2n^2$
  2. $p+1=2m^2$ & $p-1=2am^2$
  3. $p+1=2an^2$ & $p-1=2m^2$
  4. $p+1=2n^2$ & $p-1=2am^2$

Now the second and fourth of these forms give

$1=m^2-an^2$, or $1=n^2-am^2$;

which equations cannot have place, because $m$ and $n$ are less than $p$ and $q$; and these last were the least that satisfied the equation $p^2-aq^2=1$. There remain, therefore, only the first and third which give

$n^2-am^2=-1$, or $m^2-an^2=-1$

and one of these equations must necessarily obtain; but either of them resolves the equation

$x^2-ay^2=-1$

which is, therefore, always possible, when $a$ is a prime number of the form $4n+1$. It may also be observed, that, of the above two equations, the last is the only one that can obtain; for, since $n$ is odd and $m$ even, it is evident that the first cannot become equal to $-1.$

$\endgroup$
  • 2
    $\begingroup$ These arguments are due to Legendre, and there's nothing wrong with them. I have called this technique descent on Pell conics. $\endgroup$ – franz lemmermeyer Apr 12 '17 at 16:47
5
$\begingroup$

Your sentence

Of course, he was most likely wrong since we still today don't know when it is solvable

does not correctly apply, as the negative Pell question is known for prime values of $a$ in $x^2 - a y^2.$

There is a short proof in Mordell's book that for prime $p \equiv 1 \pmod 4,$ there is always a solution to $$ x^2 - p y^2 = -1. $$

The same proof does this much: if we have primes $p \equiv q \equiv 1 \pmod 4,$ while we have Legendre symbols $(p|q) = (q|p) = -1,$ there is always a solution to $$ x^2 - pq y^2 = -1. $$

When $(p|q)= (q|p) = 1$ for $p \equiv q \equiv 1 \pmod 4,$ it can go either way. The smallest examples of failure are $x^2 - 205 y^2 \neq -1$ and $x^2 - 221 y^2 \neq -1$

See both answers at Given $p \equiv q \equiv 1 \pmod 4$, $\left(\frac{p}{q}\right) = 1$, is $N(\eta) = 1$ possible?

$\endgroup$

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.