Find all positive integers $x$ such that $x(x+2021)$ is a perfect square

Question

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 $(2k-2x-2021)(2k+2x+2021)=43^2*47^2$

Answer 1

$$\begin{align}x^2+2021x-a^2&=0, a\in\mathbb Z^{+}&\end{align}$$

$$\begin{align} &\Delta =2021^2+(2a)^2=T^2, T\in\mathbb Z^{+} \\ &\implies (T-2a)(T+2a)=2021^2\\ &\implies \begin {cases} T+2a =\dfrac {2021^2}{m} \\ T-2a=m \end {cases} \\ &\implies a=\dfrac {2021^2-m^2}{4m}\end{align}$$

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.

Answer 2

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!

Answer 3

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: $$x=4,20727,22747,1020100$$

Answer 4

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, $x=ab^2\in \{ 4,1020100,22747,20727\} .$