hint

CatalinBordea wrote: Solve in $ \mathbb{Z}^2 $ the equation: $ x^2\left( 1+x^2 \right) =-1+21^y. $ This may be written $(2x^2+1)^2=4\times 21^y-3$ If $y<0$, RHS is not integer If $y=0$, this gives the solution $(x,y)=(0,0)$ If $y=1$, this gives solutions $(x,y)=(-2,1)$ and $(x,y)=(2,1)$ If $y>1$, RHS is divisible by $3$ but not by $9$, and so can not be a perfect square Hence the only solutions $\boxed{(x,y)\in\{(0,0),(-2,1),(2,1)\}}$

pco wrote: CatalinBordea wrote: Solve in $ \mathbb{Z}^2 $ the equation: $ x^2\left( 1+x^2 \right) =-1+21^y. $ If $y<0$, RHS is not integer If $y=0$, this gives the solution $(x,y)=(0,0)$ If $y=1$, this gives solutions $(x,y)=(-2,1)$ and $(x,y)=(2,1)$ If $y>1$, RHS is divisible by $3$ but not by $9$, and so can not be a perfect square Hence the only solutions $\boxed{(x,y)\in\{(0,0),(-2,1),(2,1)\}}$ The same reasoning follows after this Math-wiz wrote:

hint

Except that instead of perfect square reasoning, it will be y>2 reasoning, as y>2 will imply $9|21^y$

Multiply both sides by $4$ and add $1$ to get\[(2x^2 + 1)^2 = 4 \cdot 21^y - 3.\]Clearly $y \geq 0$ or else it happens that the RHS is not an integer. Furthermore, note that for $y \geq 2$, we have $v_3[\text{RHS}] = 1$ and since $2x^2 + 1 \equiv 0 \pmod 3$ always, we have $v_3[\text{LHS}] \geq 2$ which is a contradiction. Check that $y = 0 \implies x = 0$ and $y = 1 \implies 2x^2 + 1 = \pm 9 \implies x = \pm 2$ so our solutions are\[\boxed{(0, 0), (-2, 1), (2, 1)}\]as desired. $\blacksquare$

Is it the same to work with $\mathbb {Z}$ or with $\mathbb {Z} ^ 2 $ ?

Albert123 wrote: Is it the same to work with $\mathbb {Z}$ or with $\mathbb {Z} ^ 2 $ ? \Bump.....

Albert123 wrote: Is it the same to work with $\mathbb {Z}$ or with $\mathbb {Z} ^ 2 $ ? What do u mean? Z^2 is just a pair of integers, what exactly do you mean?

rama1728 wrote: Albert123 wrote: Is it the same to work with $\mathbb {Z}$ or with $\mathbb {Z} ^ 2 $ ? What do u mean? Z^2 is just a pair of integers, what exactly do you mean? In the initial question it says $\mathbb {Z} ^ 2 $. I was wondering if I can skip it and work with integers($\mathbb {Z}$). or $\mathbb {Z} ^ 2 $ has different properties.?

Albert123 wrote: In the initial question it says $\mathbb {Z} ^ 2 $. I was wondering if I can skip it and work with integers($\mathbb {Z}$). or $\mathbb {Z} ^ 2 $ has different properties.? Solving in $\mathbb Z$ means that we are looking for two numbers $x,y\in\mathbb Z$ Solving in $\mathbb Z^2$ means that we are looking for one pair of numbers $(x,y)\in\mathbb Z^2$ Same thing