BigSams wrote: Do there exist $\{x,y\}\in\mathbb{Z}$ satisfying $(2x+1)^{3}+1=y^{4}$? Yes. Choose for example $(x,y)=(-1,0)$

See what i got is that....by solving the equation (x+1)*( x+4-2*6^1/2 /4)*(x+4+2*^1/2 /4)=y^4 as the condition say X belongs to integer We get that only one integral solution is possible that is x= -1

sukrut wrote: See what i got is that....by solving the equation (x+1)*( x+4-2*6^1/2 /4)*(x+4+2*^1/2 /4)=y^4 as the condition say X belongs to integer We get that only one integral solution is possible that is x= -1 1) notice that your factorization is wrong 2) notice that even a good factorization would not allow you to give such immediate conclusion : Look for example at the equation $(x+1)(x+4\sqrt{14})(x-4\sqrt{14})=y^4$ The presence of square roots in LHS does not allow to conclude that the only integer solution is $(x,y)=(-1,0)$ : In this example, there is also a solution $(x,y)=(15,2)$

$(2x+1)^3=y^4-1=(y^2-1)(y^2+1)$ $(y^2-1,y^2+1)|2$ and both sides are odd $\Rightarrow (y^2-1,y^2+1)=1 \Rightarrow \exists a,b: a^3=y^2-1, b^3=y^2+1$ So $2=b^3-a^3$, which has obviously only 1 solution: $b^3=1,a^3=-1 $ Therefore $y=0$ and $x=-1$. $(-1,0)$ is indeed only solution.

sugyman im sorry but i couldn´t understand...why (y^2-1,y^2+1)|2 ?

sugyman probably meant $\gcd{(y^2-1,y^2+1)} = \gcd{(y^2-1,2)} | 2$ by Euclidean algorithm.

professordad ... exactly

We claim the pair $\boxed{(-1,0)}$ works. $\boxed{\text{ yes}}$ We have $2x+1=-1$, so $(2x+1)^3+1=0$. Also, $y^4=0^4=0$. $\blacksquare$ Not needed: To show this is the only solution, we have $(y^2-1)(y^2+1)=(2x+1)^3$. Note that $\gcd(y^2-1,y^2+1)\mid 2$, since they are odd, it is $1$. Thus, $y^2-1$ and $y^2+1$ are both perfect cubes. The only perfect cubes with difference $2$ are $-1$ and $1$, done. $\blacksquare$

Solution