Fermat, Hellegouarch, sum of powers, quadratic forms
问题内容
I am working on a sentence of Yves Hellegouarch in his book "Invitation aux mathématiques de Fermat-Wiles" . In the Fermat section, page 38, he tells that Fermat probably associated the equation $z^p=x^p+y^p$ to the form $X^2+(-1)^{(p+1)/2}pY^2$.
I don't really understand the reason he thinks so.
Here is where I am. I have for the first primes:
The "Euler" form $(u+v)^3+(u-v)^3=2u\left( u^2+3v^2\right)$ with $X^2+3Y^2$. We can guess that Fermat also made his proof with that expression
The Dirichlet form $(u+v)^5+(u-v)^5=2u\left( (u^2+5v^2)^2-5(2v)^2 \right)$ with $X^2-5Y^2$ . Fermat didn't do anything with that. It is complicated, and there aren't any traces of attempt for case $p=5$ in his correspondence.
With a lot of calculus and the help of an IA, I could get:
$(u+v)^7+(u-v)^7= 2u\left(u^2(u^2 + 7 v^2)^2 + 7v^2(u^2 - v^2)^2\right)$ with $X^2+7Y^2$
I didn't try for next primes, because it seems very complicated. I don't have knowledge about the underlying theory that could guide me. But for sure Hellegouarch knew it when he wrote his sentence. Unfortunately he does not give more explanations.
Could you tell me more about about the existence of this form $X^2+(-1)^{(p+1)/2}pY^2$, and its link with $(u+v)^p+(u-v)^p$ ? Are there explicit formulas for X and Y ? I am not sure that Fermat had the level to handle this. Indeed he worked on $a^2+nb^2$ forms, but only for some associated primes. And surely not for every n.
Thanks for your help
回答 (1)
(More of a long comment.)
For prime $p$, the OP asks about the quadratic form $X^2\pm pY^2$, more specifically $X^2+(-1)^{(p+1)/2}p Y^2$, its link with $(u+v)^p+(u-v)^p$ and asks "are there explicit formulas for X and Y"?
There probably isn't a single formula. For $p=3,5,7,11$ it is straightforward, with the latter being,
$$(u+v)^{11}+(u-v)^{11} = 2u(X^2+11Y^2)$$
$$X = u (u^4 - 22 u^2 v^2 - 11 v^4)\\ Y = v (3 u^4 - 2 u^2 v^2 - v^4)$$
However, as we go higher, there are more terms. For $p=13$, I had to use a Pell equation $p^2-13q^2 = 1$ where $(p,q)$ are half-integers like $(p,q) = \color{red}{(11/2,\, 3/2)}$ or integers like $(p,q) = \color{blue}{(649,180)}$ and, as we known, have infinitely many solutions. Thus,
$$(u+v)^{13}+(u-v)^{13} = 2u(X_1^2-13Y_1^2)$$
$$X_1 = \frac1{\color{red}2}(\color{red}{11} u^6 + 39 u^4 v^2 + 65 u^2 v^4 + 13 v^6)\\ Y_1 = \frac1{\color{red}2}(\color{red}3 u^2 + v^2)(u^2 + 3 v^2)(u^2 - v^2)$$
as well as,
$$(u+v)^{13}+(u-v)^{13} = 2u(X_2^2-13Y_2^2)$$
$$X_2 = \color{blue}{649} u^6 + 1911 u^4 v^2 + 1183 u^2 v^4 + 65 v^6\\ Y_2 = \color{blue}{180} u^6 + 530 u^4 v^2 + 328 u^2 v^4 + 18 v^6$$
and infinitely more $(X_n, Y_n)$, with coefficients getting larger and larger. So this complicates things a bit.