Congratulations, you succeed to post each of you topics in the wrong section... Please try to wait a little before posting them into the solved problems area. Pierre.

3y^2=x(x+1)(x^2-x+1) gcd(x,x+1)=1 gcd(x,x^2-x+1)=1 gcd(x+1,x^2-x+1)=1 or 3 therefore 1) or 2) or 3) is true 1) x=a^2 x+1=b^2 x^2-x+1=3c^2 obviously b>=a+1, therefore 1=b^2-a^2>=2a+1>=3 contradiction 2) x=a^2 x+1=3b^2 x^2-x+1=c^2 3) x=3a^2 x+1=b^2 x^2-x+1=c^2 (x-1)^2 < x^2-x+1 = c^2 < x^2 contradiction

2)x=a^2;x+1=3b^2; x=a^2 so x+0 or 1(mod3); x+1=3b^2,so x=2(mod3)

x=0 or 1 (mod 3) sorry!

Clearly GCD(x,y)=1 and then x is a power of 3 (x=3^L) with (L>0) odd and then 3^(3L)=A^2-1=(A-1)(A+1) A integer so A=2 and then 3L=1 Contradiction (L integer). Best regards.

This can be complicated a little to: The equation: $3z^3=x^3+y^3$ has no solution in integers except for $z=0$

Well, a simple calculation gives us that x+y=3L and easily again we get that L=1, so x+y=3., so your equation transforms into Z^3=9-3X+X^2 and X+Y=3 then if try to solve the quadratic form (in x) we get that D=4z^2-33 wich is only a square iff |D|=4 or |D|=16 then x is NOT an integer, then No integer solution if |Z|>0 hence Z=0, and Y=-X. Best regards.

Sorry for reviving this OLD problem, but can somebody explain to me why it's "clear" that gcd (x, y) = 1 ?

since $x(x^3+1)=3y^2$,$(x,x^3+1)=1$ hence $x=3t^2,x^3+1=s^2$ or $x=t^2,x^3+1=3s^2$(nod 3,contradiction!) so $x^3+1=s^2$,since $3|x$ hence $(x+1,x^2-x+1)=1$ hence $x^2-x+1=l^2$ then $x=1$,but it;s easy to check that it's not a solution.

RobertuX wrote: Well, a simple calculation gives us that $x+y=3L$ and easily again we get that $L=1$, so x+y=3., so your equation transforms into \[Z^3=9-3X+X^2\] and \[X+Y=3 \] then if try to solve the quadratic form (in x) we get that $D=4z^2-33$ wich is only a square iff |D|=4 or |D|=16 then x is NOT an integer, then No integer solution if |Z|>0 hence Z=0, and Y=-X. Best regards. Please use $\LaTeX$.

My solution: $3y^2=x^4+x=x(x^3+1)=x(x+1)(x^2-x+1)$ Because $3|3y^2 \rightarrow 3|x(x+1)(x^2-x+1)$ <1> We know that $gcd(x,x+1)=1, gcd(x,x^2-x+1)=1$ <2> Let $gcd(n+1,x^2-x+1)=d \rightarrow d|n+1 \rightarrow d|n^2+2n+1 \rightarrow d|n^2+2n+1-(n^2-n+1) \rightarrow d|3 \rightarrow d=1,3$ <3> Case 1: $3|x+1 \rightarrow 3|x^2-x+1 \rightarrow x^2-x+1=3k \rightarrow gcd(x+1,k)=1$ (from <3>) Therefore $y^2=x(x+1)k$ we see that $gcd(x,x+1)=1,gcd(x,k)=1,gcd(x+1,k)=1$ (from <1><2><3>) So $x,x+1,k$ are perfect squares so $x=a^2,x+1=b^2 \rightarrow b^2-a^2=1 \rightarrow b=1,a=0 \rightarrow x=y=0$ Case 2: $3|x \rightarrow gcd(x+1,x^2-x+1)=1$ Let $x=3q \rightarrow y^2=q(x+1)(x^2-x+1)$ We know that $gcd(q,x+1)=1,gcd(d,x^2-x+1)=1,gcd(x+1,x^2-x+1)=1$ So $q,x+1,x^2-x+1$ are perfect squares thus $x^2-x+1=c^2$ If $x=0 \rightarrow y=0$ If $x=1$ FALSE! If $x>1$ thus $(x-1)^2<x^2-x+1<x^2 \rightarrow (x-1)^2<c^2<x^2 \rightarrow x-1<c<x$ FALSE! because $c$ is an natural number. Answer: $\boxed{(x,y)=(0,0)}$

we can easily verify that x=1 is the only solution for $x^2+x+1$ to be divided by 3 using mod 3 table , and substituting x=1 , $3y^2=2 $ and we can ezly demonstrate that $sqrt\{6}$ is not an integer