Find all positive integers $x$ such that $x(x+2021)$ is a perfect square I completed the square as to get $(x+1010.5)^2-102110.25 = k^2$ but I don't know where to go from here.
Please help, thank you
I then got $(2x+2021)^2-4084441=4k^2$ then $(2k-2x-2021)(2k+2x+2021)=43^2*47^2$
 A: $$\begin{align}x^2+2021x-a^2&=0, a\in\mathbb Z^{+}&\end{align}$$
$$\begin{align} &\Delta =2021^2+(2a)^2=T^2, T\in\mathbb Z^{+} \\
&\implies (T-2a)(T+2a)=2021^2\\
&\implies \begin {cases} T+2a =\dfrac {2021^2}{m} \\ T-2a=m \end {cases} \\
&\implies a=\dfrac {2021^2-m^2}{4m}\end{align}$$
Then, note that $2021^2=43^2×47^2$ , $a \in\mathbb {Z^{+}}$ ($a \in\mathbb {Z^{+}}$ implies $T \in\mathbb {Z^{+}}$) and  $m=1,43,47,47×43, 43×47^2, 47×43^2$ .
You can check all cases.
A: For $x(x+2021)$ to be a square there has to be integers $a,b,c$ such that $$x=ab^2,x+2021=ac^2.$$
Subtracting these two equations we obtain $a(c-b)(c+b)=2021=43\times 47$. Since $43$ and $47$ are primes the only possibilities for $a, c-b$ and $c+b$ are
$$(1,43,47),(1,1,2021),(43,1,47),(47,1,43).$$
These give, in turn, $x=ab^2\in \{ 4,1020100,22747,20727\} .$
A: If the product of two relatively prime numbers is a perfect square, then both factors must themselves be perfect squares. So, let's consider that case first.
How many ways can $2021$ be expressed as a difference of two squares? The answer to that is $2$. Namely, $2021=1011^2-1010^2=45^2-2^2$. Hence, $x=2^2=4$ and $x=1010^2=1020100$ are the two solutions coprime to $2021$.
Next, suppose that $x=43n$ is divisible only by $43$ (but not by $47$), so $x+2021=43(n+47)$ must then also be divisible by $43$. Then, for $x(x+2021)$ to be a perfect square, both $n$ and $n+47$ must themselves be perfect squares, and since $47$ is prime, the only solution for $n$ is $n=23^2=529$, which leads to the solution $x=43 \cdot 529=22747$.
Likewise, suppose that $x=47n$ is divisible only by $47$ (but not by $43$), so $x+2021=47(n+43)$ must then also be divisible by $47$. Then, for $x(x+2021)$ to be a perfect square, both $n$ and $n+43$ must themselves be perfect squares, and since $43$ is prime, the only solution for $n$ is $n=21^2=441$, which leads to the solution $x=47 \cdot 441=20727$.
Finally, suppose that $x=2021n$ is divisible by $2021$ (or equivalently, by both $43$ and $47$), so $x+2021=2021(n+1)$ must then also be divisible by $2021$. Then, for $x(x+2021)$ to be a perfect square, both $n$ and $n+1$ must themselves be perfect squares. The only solution for $n$ is $n=0$, but this leads to $x=0$, which is not a positive integer.
Hence, there are exactly $4$ solutions for $x$, namely $4, 20727, 22747,$ and $1020100$. Interestingly, the number of solutions for $x$ happens to itself be one of the solutions!
A: Start from $$x (x+2021)=y^2$$
multiply both sides by $4$ and add $2021^2$ to both sides
$$4 x^2+8084 x+2021^2=4 y^2+2021^2\to (2x+2021)^2=4y^2+2021^2$$
set $2x+2021=z$ and remember that $2021^2=43^2\times 47^2$. The equation becomes
$$z^2-4y^2=43^2\times 47^2\to (z+2y)(z-2y)=43^2\times 47^2$$
we get the following possibilities
$$
\begin{array}{ll}
& \begin{cases}
z-2 y=1\\
2 y+z=2021^2\\
\end{cases}\to (y= 1021110,z= 2042221)\to x=1020100\\
&\begin{cases}
z-2 y=43\\
2 y+z=43\times 47^2\\
\end{cases}\to (y= 23736,z= 47515)\to x=22747\\
& \begin{cases}
z-2 y=47\\
2 y+z=43^2\times 47\\
\end{cases}\to (y= 21714,z= 43475)\to x=20727\\
& \begin{cases}
z-2 y=43^2\\
2 y+z= 47^2\\
\end{cases}\to (y= 90,z=  2029)\to x=4\\
\end{array}
$$
The values are only four:
$$x=4,20727,22747,1020100$$
