I have sầu seen some good proofs, but they are quite long (longer than a page) or use many variables. However, I would rather have sầu an elementary long proof with many variables than a complex short proof.
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Edit. Even if the bounty expires I will award one to someone if they have a satisfying answer.
Main idea. The proof that follows is based on the infinite descent, i.e., we shall show that if $(x,y,z)$ is a solution, then there exists another triplet $(k,l,m)$ of smaller integers, which is also a solution, và this leads apparently to lớn a contradiction.
Assume instead that $x, y, zinipes.vnbb Zsmallsetminus $ satisfy the equation (replacing $z$ by $-z$)$$x^3 + y^3 + z^3 = 0,$$ with $x, y$ and $z$ pairwise coprime. (Clearly at least one is negative.) One of them should be even, whereas the other two are odd. Assume $z$ khổng lồ be even.
Then $x$ và $y$ are odd. If $x = y$, then $2x^3 = −z^3$, & thus $x$ is also even, a contradiction. Hence $x e y$.
As $x$ & $y$ are odd, then $x+y$, $x-y$ are both even numbers. Let$$2u = x + y, quad 2v = x − y,$$where the non-zero integers $u$ và $v$ are also coprime và of different parity (one is even, the other odd), & $$x = u + vquad extandquad y = u − v.$$It follows that$$−z^3 = (u + v)^3 + (u − v)^3 = 2u(u^2 + 3v^2). ag1$$Since $u$ & $v$ have different parity, $u^2 + 3v^2$ is an odd number. And since $z$ is even, then $u$ is even & $v$ is odd. Since $u$ and $v$ are coprime, then$$ ipes.vnrmgcd,(2u,u^2 + 3v^2)=ipes.vnrmgcd,(2u,3v^2)in1,3.$$
Case I. $,ipes.vnrmgcd,(2u,u^2 + 3v^2)=1$.
In this case, the two factors of $−z^3$ in $(1)$ are coprime. This implies that $3 otmid u$ and that both the two factors are perfect cubes of two smaller numbers, $r$ and $s$.$$2u = r^3quad extandquad u^2 + 3v^2 = s^3.$$As $u^2 + 3v^2$ is odd, so is $s$. We now need the following result:
Lemma. If $ipes.vnrmgcd,(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same khung.
Proof. See here.
Thus, if $s$ is odd and if it satisfies an equation $s^3 = u^2 + 3v^2$, then it can be written in terms of two coprime integers $e$ & $f$ as$$s = e^2 + 3f^2,$$so that$$u = e ( e^2 − 9f^2) quad extandquadv = 3f ( e^2 − f^2).$$Since $u$ is even and $v$ odd, then $e$ is even & $f$ is odd. Since$$r^3 = 2u = 2e (e − 3f)(e + 3f),$$the factors $2e$, $(e–3f )$, and $(e+3f )$ are coprime since $3$ cannot divide $e$. If $3mid e$, then $3mid u$, violating the fact that $u$ and $v$ are coprime. Since the three factors on the right-hand side are coprime, they must individually equal cubes of smaller integers$$−2e = k^3,,,,e − 3f = l^3,,,,e + 3f = m^3,$$which yields a smaller solution $k^3 + l^3 + m^3= 0$. Therefore, by the argument of infinite descent, the original solution $(x, y, z)$ was impossible.
Case II. $,ipes.vnrmgcd,(2u,u^2 + 3v^2)=3$.
In this case, the greathử nghiệm comtháng divisor of $2u$ & $u^2 + 3v^2$ is $3$. That implies that $3mid u$, và one may express $u = 3w$ in terms of a smaller integer, $w$. Since $4mid u$, so is $w$; hence, $w$ is also even. Since $u$ & $v$ are coprime, so are $v$ and $w$. Therefore, neither $3$ nor $4$ divide $v$.
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Substituting $u$ by $w$ in $(1)$ we obtain$$−z^3 = 6w(9w^2 + 3v^2) = 18w(3w^2 + v^2)$$Because $v$ và $w$ are coprime, and because $3 otmid v$, then $18w$ & $3w^2 + v^2$ are also coprime. Therefore, since their hàng hóa is a cube, they are each the cube of smaller integers, $r$ & $s$:$$18w = r^3 quad extandquad3w^2 + v^2 = s^3.$$By the same lemma, as $s$ is odd và equal khổng lồ a number of the form $3w^2 + v^2$, it too can be expressed in terms of smaller coprime numbers, $e$ và $f$:$$s = e^2 + 3f^2.$$A straight-forward calculation shows that$$v = e (e^2 − 9f^2) quad extandquad w = 3f (e^2 − f^2).$$Thus, $e$ is odd & $f$ is even, because $v$ is odd. The expression for $18w$ then becomes$$r^3 = 18w = 54f (e^2 − f^2) = 54f (e + f) (e − f) = 3^3 imes 2f (e + f) (e − f).$$Since $3^3$ divides $r^3$ we have sầu that $3$ divides $r$, so $(r /3)^3$ is an integer that equals $2f (e + f) (e − f)$. Since $e$ & $f$ are coprime, so are the three factors $2e$, $e+f$, & $e−f$; therefore, they are each the cube of smaller integers, $k$, $l$, và $m$.$$−2e = k^3,,,,e + f = l^3,,,,e − f = m^3,$$which yields a smaller solution $k^3 + l^3 + m^3= 0$. Therefore, by the argument of infinite descent, the original solution $(x, y, z)$ was impossible.