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Diophantine equations are polynomial equations $F=0$, or systems of polynomial equations $F_1=\ldots=F_k=0$, where $F,F_1,\ldots,F_k$ are polynomials in either $\mathbb{Z}[X_1,\ldots,X_n]$ of $\mathbb{Q}[X_1,\ldots,X_n]$ of which it is asked to find solutions over $\mathbb{Z}$ or $\mathbb{Q}$. Topics: Pell equations, quadratic forms, elliptic curves, abelian varieties, hyperelliptic curves, Thue equations, normic forms, K3 surfaces ...
16
votes
1
answer
1k
views
Solve this Diophantine equation $(2^x-1)(3^y-1)=2z^2$
Find the positive integers $(2^x-1)(3^y-1)=2z^2$ have three solutions
$$(1,1,1),(1,2,2),(1,5,11)$$I already know $(2^x-1)(3^y-1)=z^2$ has no solution. See: P.G.Walsh December 2006 [On Diophantine equ …
14
votes
4
answers
1k
views
Find all solution $a,b,c$ with $(1-a^2)(1-b^2)(1-c^2)=8abc$
Two years ago, I made a conjecture on stackexchange:
Today, I tried to find all solutions in integers $a,b,c$ to
$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$
I have found some sol …
7
votes
Find all solution $a,b,c$ with $(1-a^2)(1-b^2)(1-c^2)=8abc$
@Allan methods it's nice! here is my answer:
since
$$\left(\dfrac{1-a^2}{2a}\right)\left(\dfrac{1-b^2}{2b}\right)\left(\dfrac{1-c^2}{2c}\right)=1$$
so let
$$\dfrac{x}{y}=\dfrac{1-a^2}{2a},\;\dfrac{y}{ …
6
votes
3
answers
846
views
Is there a simple proof that $Ax^3+By^3=C$ has only finitely many integer solutions
One can use Thue's 1909 result to show that the Diophantine equation $Ax^3 + By^3 = C$ ($A,B$ not perfect cubes, $C\neq 0$) has finitely many integer solutions $(x,y)$.
But does there exist a simple w …
4
votes
1
answer
670
views
How to solve this equation $a^2+3b^2c^2=7^c$
Let $a,b,c$ be poistive integers,and such $\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=1$,fine the all $a,b,c$ such
$$a^2+3b^2c^2=7^c$$
I'm not sure that this question has been studied, but I've been trying for a l …
4
votes
1
answer
578
views
Find all rational solutions of $x^2(x+1)(x^2+1)(x-1)=2(y+1)(y-1)$
Find all rational solutions of
$$x^2(x+1)(x^2+1)(x-1)=2(y+1)(y-1).$$
Clearly the following six solutions hold:
$$(x,y)=(1,1),(-1,-1),(-1,1),(1,-1),(0,1),(0,-1)$$
But how to find all rational solution …
3
votes
1
answer
354
views
Solve this diophantine equation: $m^4+n^4=10m^2n^2+1$
t's probably common knowledge that there are Diophantine equations which do not admit any solutions in the integers, but which admit solutions modulo nn for every nn. This fact is stated, for example, …
2
votes
2
answers
2k
views
Find all rational solutions of this diophantine-equation?
Now, today, my friend tell me this problem was posted by American Mathematical Monthly (Vol. 111, No. 2 Feb., 2004), p. 165 by Wu wei Chao ,and It is said that this problem is unsolved, until now. Fin …
2
votes
0
answers
242
views
Can estimate upper bound of $|p_{i}|$ or $|q_{i}|?$
when I Find the diophantine-equation rational points $$2y^2=x^6-x^2+2$$
I using Faltings's theorem showed that there are only finitely many solutions,if we assmue that
$(x_{i},y_{i}),i=1,2,\cdot …
0
votes
1
answer
342
views
Find the positive integers $x^3+y^3=3z^3$ [closed]
By Fermat Last theorem, I don't know if that's been discussed.
Find all positive integers $x,y,z$ such
$$x^3+y^3=3z^3$$
-1
votes
1
answer
149
views
Find the diophantine-equations $3x(x^2+2)=y^2$ integer solution [closed]
Let $x,y$ be positive integers, such that
$$3x(x^2+2)=y^2$$
since
$$3\cdot 1(1^2+2)=3\times 3=9=3^2$$
$$3\cdot 2(2^2+2)=6\cdot 6=36=6^2$$
$$24\cdot 3(24^2+2)=72\cdot 578=204^2$$
so I h …