5
$\begingroup$

When and who was the first person to discover a correct formula for the real number $r$ in terms of any given three positive distinct integers $x<y<z$ such that $$x^r + y^r = z^r\,.$$

The formula: let $$t=\log(y/x) / \log(z/x)$$ and define $f(t)$ as function of real variable $t$ and positive integer $n$ when $n$ tends to infinity:

$$f(t) = \frac{3t}{2} + (2t-1)\left(\frac{4t}{3} - 1\right) + \left(\frac{5t}{2} - 1\right)\left(\frac{5t}{3} - 1\right)\left(\frac{5t}{4} - 1\right) + \cdots + \left(\frac{nt}{2} - 1\right)\left(\frac{nt}{3} -1 \right)...\left(\frac{nt}{n-1} -1 \right)$$

Where $t$ is positive real number less than one, then: $$(x/z)^r = (\log(z/y) / \log(z/x))\cdot f(t)\,,$$or simply $r = \log \left((1-t)f(t)\right) / \log(x/z)$.

So $r$ is therefore obtainable as a function of the three positive distinct given integers $x<y<z$.

|improve this question
$\endgroup$
  • 3
    $\begingroup$ Is there indeed such a "formula"? What is it? (You could simply include a link.) $\endgroup$ – Rory Daulton Dec 24 '15 at 17:24
  • 1
    $\begingroup$ Your use of $t$ is confusing. Is it the constant defined in your second equation, or a dummy variable used to define function $f$? Which usage is it in your final equation? $\endgroup$ – Rory Daulton Dec 24 '15 at 20:25
  • 1
    $\begingroup$ I see no sources for "The formula". If you would not mind, offer a reference to a proper paper that we can all see. As Rory Daulton says, you could simply include a link. I would like to see your sourcing of this particular formula. That might bring clarity to this post. $\endgroup$ – J. W. Perry Dec 26 '15 at 5:19
  • 1
    $\begingroup$ @RoryDaulton $t$ is simply defined in terms of given integers $x, y, z$ as described above in the question, the source of this formula is unpublished book in 1994, with historical background starting from 1986, no use of showing the link, since it would show the biography only & not the content $\endgroup$ – Bassam Karzeddin Dec 26 '15 at 8:34
  • 1
    $\begingroup$ Has anyone checked the formula, (assuming there is only one counter example), then no need to go further & add unnecessary details, so please if you find one, you may kindly put it here in order to be verified, thanks $\endgroup$ – Bassam Karzeddin Dec 26 '15 at 8:43
3
$\begingroup$

Since $x^r+y^r=z^r$ can be transformed into a trinomial equation, we
could give credit to Lambert, since he developed functions for it already
in 1758, see for example this paper here by Corless, et.al.

We quickly have $x^r*(1+(y/x)^r)=x^r*(z/x)^r$.
Set $p=ln(y/x)$, $q=ln(z/x)$ and $s=e^r$. Then we get $1+s^p = s^q$.

By further setting $w=s^q$ and $t=p/q$, we get the trinomial equation
$-1+w=w^t$, which we have to solve for $w$ for a given $t$.

Edit I:
The aforementioned paper mentions a series development by Euler,
which is based on Lambert, which reads as follows:

$w^n = 1 + n*v + \frac{1}{2}*n*(n+α+β)*v^2$
$\quad \quad + \frac{1}{6}*n*(n+α+2*β)*(n+2*α+β)*v^3$
$\quad \quad + \frac{1}{24}*n*(n+α+3*β)*(n+2*α+2*β)*(n+3*α+β)*v^4$
$\quad \quad + \, etc..$

The rough form matches the one as given by the OP. But concerning
a full verification, i.e. matching the parameters α,β,v and positioning
it correctly I was too busy so far and made a side question here.

Edit II:
The equation that Euler solves is $w^α-w^β = (α-β)*v*w^{α+β}$. We can
choose $α=t-1$, $β=t$, $v=1$ and we get $w^{t-1}-w^t=(-1)*w^{2*t-1}$.

If we multiply both sides by $(-1)*w^{-t+1}$ we get $-1+w=w^t$ as required.
We can choose $n=1$ and will get a development for $w$ which is on par
with the function $f$ from the OP.

|improve this answer
$\endgroup$
  • $\begingroup$ Are you trying to understand why $r$ when $r > 2$, then it must be a transcendental number only according to current mathematics concepts? $\endgroup$ – Bassam Karzeddin Nov 6 '16 at 9:02

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.