visit "http://mathworld.wolfram.com/CatalansConjecture.html"
Does there exist any simpler or different proof of Catalans conjecture?
[Math] Any simpler proof of Catalan’s conjecture
diophantine equationsnumber theory
Related Solutions
Lemma. If $a$ and $b$ are integers such that $3a^4+3a^2+1=b^2\!$, then $a=0$ and $b=\pm 1$.
Proof. We may, without loss of generality, assume that $a$ and $b$ are non-negative. Assume, contrary to the lemma, that $a \ne 0$ and $b > 1$ are integers satisfying the equation. Evidently $b$ is odd, say $b=2c+1$ for an integer $c \ge 1$. Hence \begin{align*} 3a^4 + 3a^2+1 &= (2c+1)^2 \\ 3a^2(a^2+1) &= 4c(c+1). \end{align*} Since $4 \nmid 3(a^2+1)$, regardless of the parity of $a$, we must have $2 \mid a^2$; hence $a$ is even, say $a=2d$ for an integer $d \ge 1$. Now \begin{align*} 3d^2(4d^2+1) &= c(c+1). \end{align*} We now show that this equation has no solutions with $c,d \ge 1$. Assume to the contrary that there exist integers $p,q,r,s \ge 1$ such that \begin{align*} 3d^2 &= pq, &&& c &= pr, \\ 4d^2+1 &= rs, &&& c+1 &= qs. \end{align*} Since $\gcd(p,q) \mid \gcd(pr,qs) = \gcd(c,c+1)=1$ implies $\gcd(p,q)=1$, we may write $d=uv$ with $\gcd(u,v)=1$ and consider the two possible cases.
Case 1: $p=u^2$ and $q=3v^2$. Hence $c=pr = u^2r$, and by substitution $c+1=u^2r+1 = 3v^2s$. On the other hand, $rs-1 = 4d^2=4u^2v^2$, and adding these two relations yields \begin{align*} (rs-1)+(u^2r+1) &= 4u^2v^2 + 3v^2s \\ r(s+u^2) &= v^2(4u^2+3s). \end{align*} Since $d=uv$, the equation $rs = 4d^2+1$ forces $\gcd(v,r)=\gcd(s,u)=1$. Hence $v^2 \mid (u^2+s)$, and $\gcd(s+u^2,4u^2+3s) = \gcd(s+u^2,3(s+u^2)+u^2)=\gcd(s+u^2,u^2) = \gcd(s,u^2)=1$. Thus we conclude $r=4u^2+3s$ and $v^2=s+u^2$. Substituting now gives \begin{equation*} r = 4u^2+3s = 4u^2+3(v^2-u^2) = u^2+3v^2. \end{equation*} Multiplying, $4u^2v^2+1=rs=(u^2+3v^2)(v^2-u^2)$, which implies $u^4+1=3v^2(v^2-2u^2)$. But $3 \nmid u$ (because $3 \mid q$ and $\gcd(p,q)=1)$, so $u^4+1 \equiv 2\!\pmod{3}$, a contradiction.
Case 2: $p=3u^2$ and $q=v^2$. Now $c+1=pr+1=qs$ gives $3u^2r+1=v^2s$. We have $rs-1=4d^2=4u^2v^2$, and adding the two relations yields \begin{align*} (rs-1) + (3u^2r+1) &= 4u^2v^2 + v^2s \\ r(s+3u^2) &= v^2(4u^2+s). \end{align*} As before, considering common factors leads to the conclusion $s+3u^2=v^2$ and \begin{equation*} r=4u^2+s =4u^2+(v^2-3u^2) = u^2+v^2. \end{equation*} Multiplying, $4u^2v^2+1 = rs = (u^2+v^2)(v^2-3u^2)$, which is $3u^2(u^2+2v^2)=(v^2-1)(v^2+1)$. Since $3$ can never divide $v^2+1$, we deduce $v^2-1=3u_1^2w_1$ and $v^2+1=u_2^2w_2$ where $u=u_1u_2$ and $u^2+2v^2=w_1w_2$. As $4 \nmid (v^2+1)$, we deduce that $u_2$ is odd, $w_2 \equiv 1\text{ or }2\!\pmod{4}$, and $\gcd(u_2,w_1)=1$. Adding yields $2v^2 = (v^2-1)+(v^2+1) = 3u_1^2w_1 + u_2^2w_2$, and by substitution \begin{align*} w_1w_2 &= u^2+2v^2 \\ &= u_1^2u_2^2+ (3u_1^2w_1 + u_2^2w_2) \\ w_2(w_1-u_2^2) &= u_1^2(u_2^2+ 3w_1). \end{align*} Since $\gcd(w_1-u_2^2,u_2^2+3w_1)$ divides the sum $4w_1$, and $\gcd(w_1,u_2)=1$ and $u_2$ is odd, we deduce $w_1-u_2^2=u_1^2$. Now $w_2 = u_2^2+3w_1 = u_2^2+3(u_2^2+u_1^2) = (2u_2)^2+3u_1^2 \equiv 0\text{ or }3\!\pmod{4}$, contradicting $w_2 \equiv 1\text{ or }2\!\pmod{4}$ and proving the lemma.
Theorem. The equation $X^2 +1 = Y^3$ has only one integer solution, namely $(x,y)=(0,1)$.
Proof. Assume $x$ and $y$ are integers such that $x^2+1 = y^3$. Considering the equation modulo $3$, we quickly deduce that $$x^2 = (3z+1)^3−1 = 9z(3z^2+3z+1)$$ for some integer $z$. Since $\gcd(z,3z^2+3z+1)=1$, both $z$ and $3z^2+3z+1$ must be squares. Write $z=a^2$ for an integer $a$, and hence for some integer $b$ we have $b^2 = 3z^2+3z+1 = 3a^4+3a^2+1$. Now the lemma forces $a=0$, so that $z=a^2=0$, and $y=3z+1=1$, as claimed.
I had a little go at reading it. The paper is a mess with like $20$ help-variables together with the original $4$ defined randomly throughout making it almost impossible to follow. Finding the mistake(s) is not easy, but I think I have found a serious flaw.
The main part of the argument (his first proof) starts at page 4 (before it is just the special case $c,c' = 1$). Without using anywhere the fact that the variables have to be integers and by just defining new variables and making simple manipulations the author arrives at the result (mid-page 6) that if $Y^p = X^q + 1$ then $X^p = 4$. The only assumption is that $c = \frac{X^p - 1}{Y^{p/2}}\not= 1$ and $c' =\frac{7-X^p}{Y^{p/2}}\not= 1$. By the method used this should also hold if $X,Y,p,q$ are real numbers which is obviously wrong.
I should note that there is a possibillity that the author have been using hidden assumptions / deductions (i.e. using divisibillity properties without saying so). If this is the case then it becomes impossible to follow so at best we can say that the proof is flawed. Another issue is that the author uses cases without naming them so we have arguments like “But .. thus .. and .. hence .. thus .. or .. also .. and .. hence .. or .. and .. or .. and .. and .. hence”. What is in each case / subcase is hard to read from this. To reach the conclusion above I have tried to read the cases and sub cases as they are most naturally interpreted.
Best Answer
I'm not aware of any proof of Mihăilescu's theorem other than that by Mihăilescu himself. Searching turns up this article of Henri Cohen, but I'm not sure if it's a modification of Mihăilescu's proof or an expository account of it. (It's also in French.)
There is a nice book by René Schoof expositing the proof of Mihăilescu. It is mostly self-contained; the preface states that
(It is quite reasonable to not want to exposit class field theory for this purpose!)